multiply $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$? How is this so? I’m reading How To Prove It and in the following proof the author is doing some basic algebra with exponents that I just don’t understand.  In Step 1.) listed below he is multiplying $2^b$ across each term in (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) and gets the resulting set of terms in Step 2.)  In particular I have no idea how he is getting $2^{ab}$ from multiplying $2^{(a-1)b}$ by $2^b$ again which is shown in the first sequence in Step 2.).  When I do it I get $2^{(ab)(b) – (b)(b)}$ and assume this is as far as it can be taken.  Can someone please help me understand what steps he is taking to to get his answer?
Theorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then
$2^n$ − 1 is not prime.
Proof. Since n is not prime, there are positive integers a and b such that
a < n, b < n, and n = ab. Let x = $2^b$ − 1 and y = 1 + $2^b$ + $2^{2b}$ +· · ·+
$2^{(a−1)b}$.  Then
xy = ($2^b$ − 1) · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)
Step 1.) = $2^b$ · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) − (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)
Step 2.) = ($2^b$ + $2^{2b}$ + $2^{3b}$ +···+$2^{ab}$) − (1 + $2^b$ + $2^{2b}$ + ···+$2^{(a-1)b}$)
Step 3.) = $2^{ab}$ − 1
Step 4.) = $2^n$ − 1.

Hint:  
The correct rule is:
$$
2^b 2^{(a-1)b}=2^{(a-1)b+b}
$$
