I have a problem with following proof:
$$\gcd(n^a - 1, n^b-1) = n^{\gcd(a,b)} - 1 $$
The only thing that I can show is fact: $$n^{\gcd(a,b)} -1 | n^a - 1$$ $$n^{\gcd(a,b)} -1 | n^b - 1$$ And $k={\gcd(a,b)} $ is the biggest number that $n^k-1 | n^a - 1$ and $n^k-1|n^b-1$ But how to show that $n^{\gcd(a,b)}-1$ is $\ the\ biggest\ $ divisor or alternatively that every divisor is form $n^k-1$?
