elementary number theory - Show that $gcd(2^m-1, 2^n-1) = 2^ {gcd(m,n)} -1$

I'm trying to figure this out:

Show that for all positive integers $m$ and $n$

$\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)} -1$

I appreciate your help,

Thanks.

Note: $\gcd$ stands for the greatest common divisor.