<s>[INST] 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. 

 [/INST] Since $e_1\le e_2$, all you need is $e_1$.
$$\begin{align}e_1&=\sum_{k=1}^{25}\left\lfloor\frac{5^{25}-1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}\left\lfloor 5^{25-k}-\frac{1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}5^{25-k}+\sum_{k=1}^{25}\left\lfloor-\frac{1}{5^k}\right\rfloor\\&=\frac{5^{25}-1}{4}+25\times(-1)\\&=\frac{5^{25}-101}{4}\end{align}$$
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