can someone help me prove that gcd(p^a - 1, p^b - 1) = p^gcd(a, b) - 1, plz?

a and b are positive integers and p is a prime

this is how far i've come til i got stuck:

assume b >= a.

gcd(p^a - 1, p^b - 1) | (p^b - 1) - (p^a - 1) <=>

<=> gcd(p^a - 1, p^b - 1) | p^b - p^a <=>

<=> gcd(p^a - 1, p^b - 1) | p^a * (p^(b-a) - 1) <=>

<=> gcd(p^a - 1, p^b - 1) | p^(b-a) - 1 .. (since neither p^a - 1 or p^b - 1 divise p^a they must divise the other factor)

i don't know if i'm on the right track but it feels good, anyhow i'm stuck and would really appreciate a push (Crying)