Proof that $ p $ | ${ p^n \choose k } $ for any prime $p$ and $ k < p^n$ I know how to prove the fact that $ p$ | ${ p \choose k } $ (when writing it as a fraction, $p$ cannot be divided by any of the $1\times2\times...\times k$ or $1\times2\times...\times(p-k)$ because $p$ is prime).
When I try to apply the same rationale for $ p$ | ${ p^n \choose k } $ I get stumped because $p^n$ is in fact divisible by one of the $k! = 1\times2\times...\times k$, where $0 < k < p^n$.
If we take the example of $2^3$ and $k = 7$, we can easily see that $k! = 1 \times 2 \times ... \times 6 \times 7$, and $2$ clearly divides $2^3$.
How else can I approach this?

The highest prime power $p$ dividing $n!$ is expressed as $$\sum_{x=1}^\infty \lfloor \frac{n}{p^x} \rfloor.$$
Since $\binom{p^n}{k}=\frac{(p^n)!}{k!(p^n-k)!},$ we compare the highest prime power $p$ which divide numerator and denominator.
$$v_p((p^n)!)=\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor,$$
$$v_p(k!(p^n-k)!)=v_p(k!)+v_p((p^n-k)!)=\sum_{x=0}^n \left( \lfloor \frac{k}{p^x} \rfloor + \lfloor \frac{p^n-k}{p^x} \rfloor \right).$$
(We've replaced $\infty$ bound with $n$ since if $x>n,$ the terms evaluate to $0$).
Notice that $\lfloor a+b \rfloor \ge \lfloor a \rfloor + \lfloor b \rfloor,$ and the that if $a+b$ is an integer while $a$ and $b$ are not, then $\lfloor a+b \rfloor = \lfloor a \rfloor + \lfloor b \rfloor - 1.$ We can conclude with the first fact that $$\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor \ge \sum_{x=0}^n \left( \lfloor \frac{k}{p^x} \rfloor + \lfloor \frac{p^n-k}{p^x} \rfloor \right).$$
Because


*

*$0<k<p^n,$

*$\frac{p^n}{p^x} \in \mathbb{Z}$ for integers $1 \le x \le n,$

*$\frac{k}{p^x}+\frac{p^n-k}{p^x} \in \mathbb{Z}$ for same bounds,

*When $x=n,$ neither $\frac{k}{p^x}$ nor $\frac{p^n-k}{p^x}$ are integral $\implies \lfloor \frac{p^n}{p^n} \rfloor = 1 > \lfloor \frac{k}{p^n} \rfloor + \lfloor \frac{p^n-k}{p^n} \rfloor = 0.$
Finally, we can conclude $$\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor > \sum_{x=0}^n \left( \lfloor \frac{k}{p^x} \rfloor + \lfloor \frac{p^n-k}{p^x} \rfloor \right).$$ This implies the divisibility.
