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You can prove it using the [binomial theorem](https://en.wikipedia.org/wiki/Binomial_theorem).

Assume that $1\leq n∈\mathbb{N}$, then:

$$
\begin{align}
25^n & =(20+5)^n=\sum_{k=0}^n\binom{n}{k}20^k\cdot 5^{n-k}=5+\sum_{k=1}^{n-1}\binom{n}{k}20^k\cdot 5^{n-k}+20\\\\
 & =25+5\cdot 20\cdot\sum_{k=1}^{n-1}\binom{n}{k}20^{k-1}\cdot 5^{n-k-1}\equiv 25\mod 100
\end{align}
$$