proving via binomial theorem that $6^n-5n \equiv 1 \pmod{25}$ Use the binomial theorem to prove that remainder is always $1$ when $6^n-5n$ is divided by $25$


I haven't tried anything yet. I know only mathematical induction theorem to prove this type of question, but I can solve it with mathematical induction.
$$6^n-5n=(5+1)^n-5n=1+\sum_{k=2}^n\binom nk5^k=1+25\sum_{k=0}^{n-2}\binom n{k+2}5^k$$
$$\begin{align}
6^n-5n
&amp;= (1+5)^n -5n \\
&amp;=((n\mathop{\mathsf{C}}0)+(n\mathop{\mathsf{C}}1) \cdot 5+(n\mathop{\mathsf{C}}2) \cdot 5^2+ \cdots +(n\mathop{\mathsf{C}}n) \cdot 5^n) -5n \\
&amp;= 1+5n-5n+((n\mathop{\mathsf{C}}2) \cdot 5^2+(n\mathop{\mathsf{C}}3) \cdot 5^3 + \cdots + (n\mathop{\mathsf{C}}n) \cdot 5^n) \\
&amp;= 1+5^2((n\mathop{\mathsf{C}}2)+(n\mathop{\mathsf{C}}3)\cdot 5 +\cdots+(n\mathop{\mathsf{C}}n)5^{n-2})
\end{align}$$
Hence $1$ will always be the remainder when divided by $25.$