gcd problem

not a very strong proof (there may be a stronger one):
assuming b>a
so b=na+r
gcd(2^a-1,2^b-1)=gcd(2^a-1,2^{an+r}-1) r<a
=gcd(2^a-1,[2^an-1+1]2^r-1)
as 2^a-1 divides 2^an-1,
gcd(2^a-1,[(2^an-1)+1]2^r-1)=gcd(2^a-1,2^r-1)
proceeding as above we will reach a stage when
gcd(2^a-1,2^b-1)=gcd(2⁰-1,2^gcd(a,b)-1)=2^gcd(a,b)-1

sorry i couldnt latex the solution as i'm out of station