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There is a fairly simple proof by induction.

Base case: $n = 1$

$$25^1\equiv 25 \mod 100$$

Inductive case: Assuming $25^n\equiv 25 \mod 100$,

<p>
$$
\begin{align}
25^{n+1}&\equiv 25\cdot 25^n \mod 100\\
&\equiv 25 \cdot 25 \\
&\equiv 625 \\
&\equiv 25
\end{align}
$$
</p>

Actually, this proof can easily be extended to show the stronger statement that $5^n\equiv 25\mod 100$ for all natural numbers $n\ge2$

Base case: $n=2$

$$5^2\equiv25\mod100$$

Inductive case: Assuming $5^n\equiv 25 \mod 100$,

<p>
$$
\begin{align}
5^{n+1}&\equiv 5\cdot 5^n \mod 100\\
&\equiv 5 \cdot 25 \\
&\equiv 125 \\
&\equiv 25
\end{align}
$$
</p>

The case with powers of $25$ then simply comes from the case when $n$ is even.