Show that $a^{\gcd(m,n)}-1=\gcd(a^m-1,a^n-1)$

Question

Let $c_n$ be sequence defined for $n\ge 1$ as follows: $$ c_n=a^n-1.$$

Prove that this sequence satifies strong divisibility condition, that is: $$ \gcd(c_n, c_m)=c_{\gcd(n,m)}.$$

where $\gcd$ denotes greatest common divisor. Any hints?