Factoring $2^b-1$ out from $(1+2^a+...+2^{(b-1)a})$

Question

It is well known that $2^{ab}-1=(2^a-1)(1+2^a+...+2^{(b-1)a})=(2^b-1)(1+2^b+...+2^{(a-1)b})$. Assuming $gcd(2^a-1,2^b-1)=1$, we see $2^b-1|(1+2^a+...+2^{(b-1)a})$. My question is simply how to factor out the factor $2^b-1$ from this expression.

Answer 1

Since the primitive part of a term in a sequence is defined as the term divided by the algebraic part, and because here the algebraic part is $(2^a-1)(2^b-1)$, we know the expression we are trying to find is the primitive part of $2^{ab}-1$. This is calculable via $\Phi_{ab}(2)$, using roots of unity as shown in the link provided in the comment.