TITLE: how to prove the identity that $2^{ab}-1=(2^a-1)(2^{a(b-1)}+2^{a(b-2)}+.....+2^a+1)$
QUESTION [1 upvotes]: I am really confused.
The identity is given but I really want to know why$?$

Show that if $2^n -1$ is a prime , then $n$ is a prime . [Hint : Use the identity] 
$$2^{ab} -1 = (2^a-1)\cdot(2^{a(b-1)} + 2^{a(b-2)} + \cdots + 2^a +1 )$$

REPLY [2 votes]: Consider the following G.P :
$$1,2^a , 2^{2a} ,\cdots ,2^{a(b-2)} , 2^{a(b-1)} $$
We use the Formula to find it's sum :
$$S = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}$$
Or
$${\color{#d05}{1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)} = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}}}$$
Rearranging this gives us the deisred result :
$${\color{#20f}{({2^a-1})(1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)}) = {(2^{ab}-1)}}}$$
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