Proving $\gcd(N^a-1,N^b-1)=N^{\gcd(a,b)}-1$.

Question

I have come by one solution only, but things were derived too quickly without me understanding how or why. How does knowing that $\gcd(a,b)$ is a factor and a and b, actually derive that $N^{\gcd(a,b)}-1$ is a factor of $N^a-1$ and $N^b-1$? How is it derived? I tried using congruence but it wouldn't work, and I can't seem to divide things that have summation and subtraction in them, Let alone show $N^{\gcd(a,b)}-1$ is the greatest. I could really use any help.

Answer 1

Hint: WLoG suppose that $a\le b$, then we have $$ N^b-1=N^aN^{b-a}-1=(N^a-1+1)N^{b-a}-1=(N^a-1)N^{b-a}+N^{b-a}-1 $$ Now use Euclid's algorithm to find that $$ (N^a-1,N^b-1)=(N^a-1,N^{b-a}-1) $$