# Infinite product simplification

I found the following identity $\prod_{k=0}^{\infty}(1+\frac{1}{2^{2^k}-1})=\frac{1}{2}+\sum_{k=0}^{\infty}\frac{1}{\prod_{j=0}^{k-1}(2^{2^j}-1)}$

My first thought was to use eulers identity somehow $\prod_{k=1}^{\infty}(1+z^k)=\prod_{k=1}^{\infty}(1-z^{2k-1})^{-1}$ but it does not help me.

If you have an idea or know a helpful identity to prove this result I would really appreciate it.

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My first thought was to "expand the terms of the infinite product." – Lord Soth Apr 2 '13 at 19:47
But, then I said to myself "Well, that may not be a good idea, greed for getting the second term is not cool." – Lord Soth Apr 2 '13 at 20:00
Maybe using a recursive formula for $2^{2^n}-1$: $a(n)=(a(n-1)+1)^2-1, a(0)=1$ – Babla Apr 2 '13 at 20:04
My first reaction is to use the geometric series $1/(T-1) = 1/T + 1/T^2 + 1/T^3 + \cdots$ on both sides of the equation (with $T=2^{2^k}$ and $T=2^{2^j}$) and see if that helps. – Greg Martin Apr 2 '13 at 20:04
The funny thing is that while working on this problem, I "discovered" an extremely fast converging series that can approximate the constant $1$: $\sum_{k=0}^{\infty}\frac{2^{2^k}-1}{2^{2^{k+1}-1}} = 1$. – Lord Soth Apr 2 '13 at 20:29