Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.

$\color{#c00}{a\mid 1\!+\!b,\ n\geq 2}\,\Rightarrow\, (\color{#c00}a\,\color{#0a0}{b})^{\large 2}\!\mid (1\!+\!b)^{\large n+b}\!-{(1\!+\!b)^{\large n}}  = \overbrace{(\color{#c00}{1\!+\!b})^{\large\color{#c00} n}}^{\Large \color{#c00}{a^{\Large 2}}\,(\cdots)}\,\underbrace{\overbrace{((1\!+\!b)^{\large b} - 1)}^{\Large \color{#0a0}{b^{\Large 2}}(\cdots)}}_{\rm{ Binomial\  Theorem}}.\ $ Put $\,a=2,\,b=5$.
