text: Highest power of a prime $p$ dividing $N!$
  
  How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
  
  Related question: How many zeros are there at the end of $N!$?
  
  This is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details.
  
  10 Answers
  
  Largest power of a prime dividing $N!$
  
  In general, the highest power of a prime $p$ dividing $N!$ is given by
  
  $$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
  
  The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.
  
  Number of zeros at the end of $N!$
  
  The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.
  
  To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.
  
  Note that $\alpha_5 < \alpha_2$ whenever $N \geq 2$. (Why?)
  
  The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$
  
  If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.
  
  Note that there will be
  
  1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \mathrel\| N$
  
  2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \mathrel\| N$
  
  3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \mathrel\| N$ and in general
  
  4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \mathrel\| N$
  
  where $a \mathrel\| b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1.
  
  Largest power of a prime dividing other related products
  
  In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.
  
  For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.
  
  Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.
  
  Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above.
  
  • In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. Commented May 5, 2012 at 1:32
  • @AndréNicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$?
  – user17762
  Commented May 5, 2012 at 1:36
  • One trick that makes computations easier is $$\left\lfloor \dfrac{N}{p^{n+1}} \right\rfloor = \left\lfloor \dfrac{\left\lfloor \dfrac{N}{p^n} \right\rfloor}{p} \right\rfloor$$ Commented Jun 1, 2016 at 8:26
  
  For a number $n$, define $\Lambda(n)=\log p$ if $n=p^k$ and zero elsewhen.
  
  Note that $\log n=\sum\limits_{d\mid n}\Lambda(d)$. If $N=n!$, then $$\log N=\sum_{k=1}^n\log k=\sum_{k=1}^n\sum_{d\mid k}\Lambda (d)=\sum_{d=1}^n \Lambda(d)\left\lfloor \frac nd\right\rfloor$$
  
  Since $\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\log p$, the last sum equals $$\sum_{p}\sum_{k\geqslant 1}\log p \left\lfloor\frac n{p^k}\right\rfloor$$ and this gives $$\nu_p(n!)=\sum_{k\geqslant 1}\left\lfloor\frac n{p^k}\right\rfloor$$ If you write $n$ is base $p$, say $n=a_0+a_1p+\cdots+a_kp^k$, the above gives that $$\nu_p(n!)=\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\cdots+a_k$.
  
  • Very nice proof with using von Mangoldt function! +1
  – RFZ
  Commented Jun 9, 2016 at 11:08
  • Cool and simple. I think the hardest part is the last equality in the first formula. But it becomes obvious if you change the summation order. Commented Jun 6, 2020 at 10:18
  
  de Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$.
  
  • Sometimes the formula is also attributed to Legendre. Commented May 5, 2012 at 4:46
  
  For something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write:
  
  (n-Total[IntegerDigits[n,p]])/(p-1)
  
  
  Hence, the number of trailing zeros of $n!$ is
  
  (n-Total[IntegerDigits[n,5]])/4
  
  
  I've found this method to be much faster than the sum of Floor functions...
  
  Walking down the street I realized how to proof the formula in a very simple way. The key idea is to know how many numbers contain a given prime in the exact $$q$$-th power.
  
  Let us denote the set of all numbers up to $$n$$ (inclusive) that are divisible by $$q$$-th power of prime $$p$$ but not divisible by the greater ($$q+1, q+2, ...$$) power of that prime: $$M^p_q(n) := \{ m : m \le n, p^q | m, p^{q+1} \nmid m\}$$ How many numbers in that set? Well, obviously, the number of those that are divisible by $$p^q$$ minus the number of those which are divisible by $$p^{q+1}$$: $$|M^p_q(n)| = \bigg\lfloor \frac{n}{p^q}\bigg\rfloor - \bigg\lfloor \frac{n}{p^{q+1}}\bigg\rfloor$$ OK. Each of the numbers in $$M^p_q$$ gives us the $$q$$-th power of $$p$$ in $$n!$$ so we need just add them all: $$s_p(n!)=\sum_{q=1}^{\infty}q|M^p_q(n)|$$ Which is the desired result.
  
  Define $e_p(n) = s_p(n!),$ as Marvis is using $s_p$ for the highest exponent function...
  
  We get $$e_p(n) = s_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor$$
  
  I thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$
  
  LEMMA:
  
  (A) If $n + 1 \equiv 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = 1 + \left\lfloor \frac{n}{m} \right\rfloor$$ (B) If $n + 1 \neq 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = \left\lfloor \frac{n}{m} \right\rfloor$$
  
  For $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ are $0$ as well. So the base cases of the induction is true.
  
  Now for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change.
  
  If $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction.
  
  Note that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \geq p,$ because we are not factoring $n,$ we are factoring $n!$
  
  Here's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $$p$$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it.
  
  Let's consider $$s_p(n)$$, the sum of digits of $$n$$ in base $$p$$. How does the sum of digits change if we add $$1$$ to it? It usually just increments the last digit by 1, so most of the time,
  
  $$s_p(n+1) = s_p(n)+1$$
  
  But that's not always true. If the last digit was $$p-1$$ we'd end up carrying, so we'd lose $$p-1$$ in the sum of digits, but still gain $$1$$. So the formula in that case would be,
  
  $$s_p(n+1) = s_p(n) + 1 - (p-1)$$
  
  But what if after carrying, we end up carrying again? Then we'd keep losing another $$p-1$$ term. Every time we carry, we leave a $$0$$ behind as a digit in that place, so the total number of times we lose $$p-1$$ is exactly the number of $$0$$s that $$n+1$$ ends in, $$v_p(n+1)$$.
  
  $$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$
  
  Now let's rearrange to make the telescoping series and sum over,
  
  $$\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$ $$s_p(k) = k - (p-1) \sum_{n=0}^{k-1}v_p(n+1)$$
  
  Notice that since $$v_p$$ is completely additive so long as $$p$$ is a prime number,
  
  $$s_p(k) = k - (p-1) v_p\left(\prod_{n=0}^{k-1}(n+1)\right)$$
  
  $$s_p(k) = k - (p-1) v_p(k!)$$
  
  This rearranges to the well-known Legendre's formula,
  
  $$v_p(k!) = \frac{k-s_p(k)}{p-1}$$
  
  • WOW! Not bad. Though I would recommend you to add the parentheses in order to show what are the terms in the summations. It's a bit confusing. BTW. It's funny that you gave the answer to this old question almost at the same time as I did. Commented Jun 6, 2020 at 10:31
  
  $$s_{p}(n!)=\sum_{i=1}^{\infty}[\frac{n}{p^i}]$$
  
  Since $$n!$$ is the product of the integers $$1$$ through $$n$$, we have at least one factor of $$p$$ in $$n!$$ for each multiple of $$p$$ in $$\{1,2,\ldots ,n\}$$, of which there are $$[ \frac{n}{p}]$$. Each multiple of $$p^2$$ contributes an additional factor of $$p$$, each multiple of $$p^3$$ contributes yet another factor of $$p$$, etc. Adding up the number of these factors gives the infinite sum for $$s _{p}(n!)$$.
  
  This is essentially a variation on the answer by LRDPRDX.
  
  For each $$k\geq 1$$ let $$M_k$$ denote the set of positive multiples of $$p^k$$, i.e, $$M_k=\{a p^k: a\geq 1 \text{ and } ap^k\leq n\}.$$ Observe that $$M_{k+1}\subset M_k$$, $$M_k=\emptyset$$ for $$k>\lfloor n/(\log p)\rfloor$$ and $$|M_k| = \lfloor p/n^k\rfloor$$.
  
  The $$p$$-adic valuation of $$n!$$ is
  
  \begin{align} &\sum_{i=2}^n \nu_p(i) = \sum_{k=1}^\infty k |\{i\in \{2,\ldots, n\}: \nu_p(i) = k\}| \\&=\sum_{k=1}^\infty k |M_k \setminus M_{k+1}| \\&= \sum_{k=1}^\infty k (|M_k| - |M_{k+1}|) \\&= \sum_{k=1}^\infty k |M_k| - \sum_{k=2}^\infty (k-1)|M_{k}| \\&= \sum_{k=1}^\infty k |M_k| - \sum_{k=1}^\infty (k-1)|M_{k}| \\&= \sum_{k=1}^\infty |M_k| = \sum_{k=1}^\infty \lfloor p/n^k\rfloor. \end{align}
  
  Maximum value of x such that $$\frac{n!}{p^x}\in\mathbb{Z}$$:
  
  $$n!=n(n-1)(n-2)\ldots{}2=n_p q$$, where $$p\mid n_p \land p\nmid q$$ . Therefore: $$\frac{n!}{p^x}\in\mathbb{Z} \Leftrightarrow \frac{n_p}{p^x}\in\mathbb{Z}$$ $$n_p = \prod_{i=1}^kip, \,k = \lfloor \frac{n}{p}\rfloor$$ $$\Updownarrow$$ $$n_p = p^k\prod_{i=1}^ki = p^k k!$$ Let $$\gamma(n!)_p$$ be the function which returns the max value of x. Then: $$\gamma(n!)_p = \lfloor\frac{n}{p}\rfloor + \gamma(\lfloor\frac{n}{p}\rfloor !)_p$$ $$\gamma(0)_p = 0$$ $$\gamma(n)_p$$ can also be interpreted as the sum: $$\sum_{i=1}^m\lfloor \frac{n}{p^i}\rfloor$$ Where m is the largest value of $$x$$ where $$\lfloor\frac{n}{p^x}\rfloor>0$$.
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distilabel_metadata.content: Highest power of a prime $p$ dividing $N!$
  
  How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
  
  Related question: How many zeros are there at the end of $N!$?
  
  This is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details.
  
  10 Answers
  
  Largest power of a prime dividing $N!$
  
  In general, the highest power of a prime $p$ dividing $N!$ is given by
  
  $$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
  
  The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.
  
  Number of zeros at the end of $N!$
  
  The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.
  
  To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.
  
  Note that $\alpha_5 < \alpha_2$ whenever $N \geq 2$. (Why?)
  
  The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$
  
  If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.
  
  Note that there will be
  
  1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \mathrel\| N$
  
  2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \mathrel\| N$
  
  3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \mathrel\| N$ and in general
  
  4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \mathrel\| N$
  
  where $a \mathrel\| b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1.
  
  Largest power of a prime dividing other related products
  
  In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.
  
  For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.
  
  Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.
  
  Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above.
  
  • In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. Commented May 5, 2012 at 1:32
  • @AndréNicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$?
  – user17762
  Commented May 5, 2012 at 1:36
  • One trick that makes computations easier is $$\left\lfloor \dfrac{N}{p^{n+1}} \right\rfloor = \left\lfloor \dfrac{\left\lfloor \dfrac{N}{p^n} \right\rfloor}{p} \right\rfloor$$ Commented Jun 1, 2016 at 8:26
  
  For a number $n$, define $\Lambda(n)=\log p$ if $n=p^k$ and zero elsewhen.
  
  Note that $\log n=\sum\limits_{d\mid n}\Lambda(d)$. If $N=n!$, then $$\log N=\sum_{k=1}^n\log k=\sum_{k=1}^n\sum_{d\mid k}\Lambda (d)=\sum_{d=1}^n \Lambda(d)\left\lfloor \frac nd\right\rfloor$$
  
  Since $\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\log p$, the last sum equals $$\sum_{p}\sum_{k\geqslant 1}\log p \left\lfloor\frac n{p^k}\right\rfloor$$ and this gives $$\nu_p(n!)=\sum_{k\geqslant 1}\left\lfloor\frac n{p^k}\right\rfloor$$ If you write $n$ is base $p$, say $n=a_0+a_1p+\cdots+a_kp^k$, the above gives that $$\nu_p(n!)=\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\cdots+a_k$.
  
  • Very nice proof with using von Mangoldt function! +1
  – RFZ
  Commented Jun 9, 2016 at 11:08
  • Cool and simple. I think the hardest part is the last equality in the first formula. But it becomes obvious if you change the summation order. Commented Jun 6, 2020 at 10:18
  
  de Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$.
  
  • Sometimes the formula is also attributed to Legendre. Commented May 5, 2012 at 4:46
  
  For something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write:
  
  (n-Total[IntegerDigits[n,p]])/(p-1)
  
  
  Hence, the number of trailing zeros of $n!$ is
  
  (n-Total[IntegerDigits[n,5]])/4
  
  
  I've found this method to be much faster than the sum of Floor functions...
  
  Walking down the street I realized how to proof the formula in a very simple way. The key idea is to know how many numbers contain a given prime in the exact $$q$$-th power.
  
  Let us denote the set of all numbers up to $$n$$ (inclusive) that are divisible by $$q$$-th power of prime $$p$$ but not divisible by the greater ($$q+1, q+2, ...$$) power of that prime: $$M^p_q(n) := \{ m : m \le n, p^q | m, p^{q+1} \nmid m\}$$ How many numbers in that set? Well, obviously, the number of those that are divisible by $$p^q$$ minus the number of those which are divisible by $$p^{q+1}$$: $$|M^p_q(n)| = \bigg\lfloor \frac{n}{p^q}\bigg\rfloor - \bigg\lfloor \frac{n}{p^{q+1}}\bigg\rfloor$$ OK. Each of the numbers in $$M^p_q$$ gives us the $$q$$-th power of $$p$$ in $$n!$$ so we need just add them all: $$s_p(n!)=\sum_{q=1}^{\infty}q|M^p_q(n)|$$ Which is the desired result.
  
  Define $e_p(n) = s_p(n!),$ as Marvis is using $s_p$ for the highest exponent function...
  
  We get $$e_p(n) = s_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor$$
  
  I thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$
  
  LEMMA:
  
  (A) If $n + 1 \equiv 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = 1 + \left\lfloor \frac{n}{m} \right\rfloor$$ (B) If $n + 1 \neq 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = \left\lfloor \frac{n}{m} \right\rfloor$$
  
  For $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ are $0$ as well. So the base cases of the induction is true.
  
  Now for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change.
  
  If $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction.
  
  Note that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \geq p,$ because we are not factoring $n,$ we are factoring $n!$
  
  Here's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $$p$$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it.
  
  Let's consider $$s_p(n)$$, the sum of digits of $$n$$ in base $$p$$. How does the sum of digits change if we add $$1$$ to it? It usually just increments the last digit by 1, so most of the time,
  
  $$s_p(n+1) = s_p(n)+1$$
  
  But that's not always true. If the last digit was $$p-1$$ we'd end up carrying, so we'd lose $$p-1$$ in the sum of digits, but still gain $$1$$. So the formula in that case would be,
  
  $$s_p(n+1) = s_p(n) + 1 - (p-1)$$
  
  But what if after carrying, we end up carrying again? Then we'd keep losing another $$p-1$$ term. Every time we carry, we leave a $$0$$ behind as a digit in that place, so the total number of times we lose $$p-1$$ is exactly the number of $$0$$s that $$n+1$$ ends in, $$v_p(n+1)$$.
  
  $$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$
  
  Now let's rearrange to make the telescoping series and sum over,
  
  $$\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$ $$s_p(k) = k - (p-1) \sum_{n=0}^{k-1}v_p(n+1)$$
  
  Notice that since $$v_p$$ is completely additive so long as $$p$$ is a prime number,
  
  $$s_p(k) = k - (p-1) v_p\left(\prod_{n=0}^{k-1}(n+1)\right)$$
  
  $$s_p(k) = k - (p-1) v_p(k!)$$
  
  This rearranges to the well-known Legendre's formula,
  
  $$v_p(k!) = \frac{k-s_p(k)}{p-1}$$
  
  • WOW! Not bad. Though I would recommend you to add the parentheses in order to show what are the terms in the summations. It's a bit confusing. BTW. It's funny that you gave the answer to this old question almost at the same time as I did. Commented Jun 6, 2020 at 10:31
  
  $$s_{p}(n!)=\sum_{i=1}^{\infty}[\frac{n}{p^i}]$$
  
  Since $$n!$$ is the product of the integers $$1$$ through $$n$$, we have at least one factor of $$p$$ in $$n!$$ for each multiple of $$p$$ in $$\{1,2,\ldots ,n\}$$, of which there are $$[ \frac{n}{p}]$$. Each multiple of $$p^2$$ contributes an additional factor of $$p$$, each multiple of $$p^3$$ contributes yet another factor of $$p$$, etc. Adding up the number of these factors gives the infinite sum for $$s _{p}(n!)$$.
  
  This is essentially a variation on the answer by LRDPRDX.
  
  For each $$k\geq 1$$ let $$M_k$$ denote the set of positive multiples of $$p^k$$, i.e, $$M_k=\{a p^k: a\geq 1 \text{ and } ap^k\leq n\}.$$ Observe that $$M_{k+1}\subset M_k$$, $$M_k=\emptyset$$ for $$k>\lfloor n/(\log p)\rfloor$$ and $$|M_k| = \lfloor p/n^k\rfloor$$.
  
  The $$p$$-adic valuation of $$n!$$ is
  
  \begin{align} &\sum_{i=2}^n \nu_p(i) = \sum_{k=1}^\infty k |\{i\in \{2,\ldots, n\}: \nu_p(i) = k\}| \\&=\sum_{k=1}^\infty k |M_k \setminus M_{k+1}| \\&= \sum_{k=1}^\infty k (|M_k| - |M_{k+1}|) \\&= \sum_{k=1}^\infty k |M_k| - \sum_{k=2}^\infty (k-1)|M_{k}| \\&= \sum_{k=1}^\infty k |M_k| - \sum_{k=1}^\infty (k-1)|M_{k}| \\&= \sum_{k=1}^\infty |M_k| = \sum_{k=1}^\infty \lfloor p/n^k\rfloor. \end{align}
  
  Maximum value of x such that $$\frac{n!}{p^x}\in\mathbb{Z}$$:
  
  $$n!=n(n-1)(n-2)\ldots{}2=n_p q$$, where $$p\mid n_p \land p\nmid q$$ . Therefore: $$\frac{n!}{p^x}\in\mathbb{Z} \Leftrightarrow \frac{n_p}{p^x}\in\mathbb{Z}$$ $$n_p = \prod_{i=1}^kip, \,k = \lfloor \frac{n}{p}\rfloor$$ $$\Updownarrow$$ $$n_p = p^k\prod_{i=1}^ki = p^k k!$$ Let $$\gamma(n!)_p$$ be the function which returns the max value of x. Then: $$\gamma(n!)_p = \lfloor\frac{n}{p}\rfloor + \gamma(\lfloor\frac{n}{p}\rfloor !)_p$$ $$\gamma(0)_p = 0$$ $$\gamma(n)_p$$ can also be interpreted as the sum: $$\sum_{i=1}^m\lfloor \frac{n}{p^i}\rfloor$$ Where m is the largest value of $$x$$ where $$\lfloor\frac{n}{p^x}\rfloor>0$$.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed exploration of the problem of finding the highest power of a prime \( p \) that divides \( N! \) and the number of trailing zeros in \( N! \). The text contains multiple questions and their corresponding answers, making it a comprehensive resource on the topic.
  
  1. **Highest power of a prime \( p \) dividing \( N! \)**:
     - **Question**: How does one find the highest power of a prime \( p \) that divides \( N! \) and other related products?
     - **Answer**: The highest power of a prime \( p \) dividing \( N! \) is given by:
       \[
       s_p(N!) = \left\lfloor \frac{N}{p} \right\rfloor + \left\lfloor \frac{N}{p^2} \right\rfloor + \left\lfloor \frac{N}{p^3} \right\rfloor + \cdots
       \]
       This formula accounts for the number of multiples of \( p \), \( p^2 \), \( p^3 \), etc., up to \( N \).
  
  2. **Number of zeros at the end of \( N!