How to prove that $\gcd((a^m -1)/(a-1),a-1)=\gcd((a-1),m)$

Question

I have this identity $$ \gcd\left({a^m -1\over a-1},a-1\right)=\gcd(a-1,m) $$ where

• $m$ is positive integer and
• $a>1$.

How can I prove it?

I tried to do so by expanding the expression $(a^m - 1 )/ (a-1)$ and then use the Euclidean algorithm or use the $\gcd$ properties, but still I wasn't able to prove it.