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Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$.

 > Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$.

This was a simple observation I made when playing around and I came up with the following proof:


 > It follows from $(10a + 25)^2 = 100a^2 + 500a + 625 = 100(a^2 + 5a) + 625$ that any number ending in $25$ raised to a power of $2$ will also end in $25$. As each number has a unique binary representation, every number can be written as a sum of powers of two. By the rule $a^{b + c} = a^b \cdot a^c$ and $ab \bmod {c} = (a \bmod {c}) \cdot (b \bmod {c}) \bmod {c}$, $25^n \mod {100}$ will eventually cascade down to $25 \bmod {100}$. For example, \begin{align}25^{19} \equiv 25^{16 + 2 + 1} \equiv 25^{16} \cdot 25^{2} \cdot 25^{1} \equiv 25 \cdot 25 \cdot 25 \\\\ \equiv 25^{3} \equiv 25^{2 + 1} \equiv 25^2 \cdot 25^1 \equiv 25 \cdot 25 \equiv 25^2 \equiv 25\bmod{100}\end{align}

I am wondering if there are alternate, potentially simpler proofs.