($2^{(2^0)}$ + 1) * ($2^{(2^1)}$ + 1)* ... *($2^{(2^k)}$ + 1) = $2^{(2^{(k+1)})} - 1$
I've seen this is true for k = 1,2,3,4
All help appreciated. I've never worked with an exponent to an exponent before, so I don't know how to apply rules of a finite geometric series.
It can be easily shown with induction :
let $$F(n) = \prod_{k=0}^{n-1} \left( 2^{(2^k)} + 1 \right).$$ If $n = 1$, it is trivial that $F(1) = 2^{(2^0)} + 1 = 2^{(2^1)} - 1 = 3.$
For $n \in \mathbb{N}$, suppose that $$F(n) = 2^{(2^n)} - 1,$$ and multiply $2^{(2^{n})} + 1$ on both sides : $$(2^{(2^{n})} + 1) F(n) = F(n+1) = (2^{(2^{n})} + 1)(2^{(2^n)} - 1) = 2^{(2^{n+1}\:)} - 1$$ and the statement is proved.