Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$ 
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$

Attempt:
I want to use the following theorem:

The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$e=[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+\cdots$$.

The number of times the prime divisor $5$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $5$ contained in $(5^{25}-1)!$, which is  $e_1=[\frac{(5^{25}-1)}{5}]+[\frac{(5^{25}-1)}{5^2}]+[\frac{(5^{25}-1)}{5^3}]+[\frac{(5^{25}-1)}{5^4}]+\cdots$
The number of times the prime divisor $2$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $2$ contained in $(5^{25}-1)!$, which is $e_2=[\frac{(5^{25}-1)}{2}]+[\frac{(5^{25}-1)}{2^2}]+[\frac{(5^{25}-1)}{2^3}+[\frac{(5^{25}-1)}{2^4}]+\cdots$
Therefore, the number of zeros at the right end equals the greatest exponent of $10$ contained in $(5^{25}-1)!$$=\min{(e_1, e_2)}$. 

I am unable to simplify $e_1, e_2$. Please help me. 


As shown in this answer, the number of factors of a prime $p$ in $n!$ is
$$
\frac{n-\sigma_p(n)}{p-1}
$$
where $\sigma_p(n)$ is the sum of the digits in the base-$p$ expansion of $n$.

The base-$5$ expansion of $5^{25}-1$ consists of $25\,\,4$s, thus
$$
\sigma_{5}(5^{25}-1)=25\cdot4=100
$$
Thus, the number of factors of $5$ in $(5^{25}-1)!$ is
$$
\frac{5^{25}-1-100}{5-1}=\frac{5^{25}-101}{4}
$$

The binary representation of $5^{25}-1$ has at most $3\cdot25=75$ bits since $5\lt2^3$. Thus, the number of factors of $2$ in $(5^{25}-1)!$ is at least
$$
\frac{5^{25}-1-75}{2-1}=5^{25}-76\gt\frac{5^{25}-101}{4}
$$

Thus, the number of factors of $10$ in $(5^{25}-1)!$ is $\dfrac{5^{25}-101}{4}$.
