Determine the largest power of 10 that is a factor of $50!\,$? How would one find the largest power of 10 which is a factor of $50!\,{}$?

This problem is equivalent to find the number of trailing zeros in $n!=50!$. I adapt this old answer of mine to the question Derive a formula to find the number of trailing zeroes in $n!$.
For every integer $n$ the exponent of the prime $p$ in the prime
factorization of $n!$ is given by de Polignac's formula $$\displaystyle\sum_{i\ge 1}\left\lfloor \frac{n}{p^{i}}
\right\rfloor =\left\lfloor \frac{n}{p}\right\rfloor+\left\lfloor \frac{n}{p^2}\right\rfloor+\cdots+\left\lfloor \frac{n}{p^k}\right\rfloor ,\qquad p^k\le n,\ p^{k+1}\gt n.$$
because this exponent is obtained by adding to the numbers between $1$ and $n$ which are
divisible by $p$ the number of those divisible by $p^{2}$, then the number
of those divisible by $p^{3}$, and so on. The process terminates at the
greatest power $p^{i}\leq n$.
For $n=50$ the exponent of $5$ in in the prime factorization of $50!$ is  $$e_5(50!)=\sum_{i\geq 1}\left\lfloor \dfrac{50}{5^{i}}\right\rfloor=\left\lfloor \dfrac{50}{5}\right\rfloor +\left\lfloor \dfrac{50}{5^{2}}\right\rfloor =10+2=12<47.$$
Similarly the exponent of $2$ in the prime factorization of $50!$ is
$$\begin{align}
e_2(50!)=\sum_{i\geq 1}\left\lfloor \dfrac{50}{2^{i}}\right\rfloor &= \left\lfloor \dfrac{50}{2}\right\rfloor+\left\lfloor \dfrac{50}{2^{2}} \right\rfloor + \left\lfloor \dfrac{50}{2^{3}} \right\rfloor + \left\lfloor\dfrac{50}{2^{4}}\right\rfloor +\left\lfloor \dfrac{50}{2^{5}}\right\rfloor
\\
&=25+12+6+3+1 
\\
&=47>12.
\end{align}$$
So $50!=2^{47}5^{12}m=(2^{35}m)10^{12}$, with $\gcd(m,10)=1$, as pointed out by robjohn. Consequently, the number of trailing zeroes in $50!$ is $12$ and the power is $10^{12}$.
