# Prove that $2^{pq} - 1 = (2^p - 1)(\sum^{q-1}_{i=0} 2^{pi})$ for two natural numbers $p, q$

How can I prove the below?

$$2^{pq} - 1 = (2^p - 1)\left(\sum^{q-1}_{i=0} 2^{pi}\right)$$ for two natural numbers $p, q$

It looks like I need proof by induction? But how? There's two variables?

-

Setting $a=2^p$ then $$2^{pq}-1=a^q-1=(a-1)\sum_{i=0}^{q-1}a^i=(2^p-1)\sum_{i=0}^{q-1}(2^p)^i=(2^p-1)\sum_{i=0}^{q-1}2^{pi}.$$