The remainder when $6^n – 5^n$ is divided by $25$, is

$\begin{array}{l}{6}^n-5n=(1+5{)}^n-5n\\ =1+5n{+}^n{C}_2\left(5^2\right){+}^n{C}_3\left(5^3\right)+\dots {+}^n{C}_n\left(5^n\right)\\ -5n\\ =1+25m\end{array}$

Where  $m{=}^n{C}_2{+}^n{C}_3\left(5\right)+\dots {+}^n{C}_n\left(5^{n-2}\right)$ is a whole number.
Thus, remainder when $6^n – 5^n$ is divided by $25$ is $1.$