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. 
 A: To prove this identity, observe that the left-hand side is
\begin{eqnarray*}
\prod_{k\ge 0} (1+\frac{1}{2^{2^k}-1})
&=&
\prod_{k\ge 0} \frac{2^{2^k}}{2^{2^k}-1}\\
&=& \prod_{k\ge 0} \frac{1}{1-2^{-2^k}}\\
&=& \prod_{k\ge 0} (1 + 2^{-2^k} + 2^{-2\cdot 2^k} + 2^{-3\cdot 2^k} + \cdots)
\end{eqnarray*}
and then, partially expanding the infinite product, 
\begin{eqnarray*}
&\ & \prod_{k\ge 0} (1 + 2^{-2^k} + 2^{-2\cdot 2^k} + 2^{-3\cdot 2^k} + \cdots)\\
&=& 1+\sum_{k\ge 0} (2^{-2^k} + 2^{-2\cdot 2^k}+ 2^{-3\cdot 2^k} + \cdots)\prod_{0\le j<k}  (1 + 2^{-2^j} + 2^{-2\cdot 2^j} + 2^{-3\cdot 2^j} + \cdots)\\
&=& 1+\sum_{k\ge 0} \left(2^{-2^k} + \frac{2^{-2\cdot 2^k}}{1-2^{-2^k}}\right)
\prod_{0\le j<k} (1-2^{-2^j})^{-1}\\
&=& 1+\sum_{k\ge 0} \left(2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}+ 2^{-2^{k+1}} \prod_{0\le j\le k} (1-2^{-2^j})^{-1}\right)\\
&=& 1+\sum_{k\ge 0} 2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}
+\sum_{\ell\ge 1} 2^{-2^\ell} \prod_{0\le j<\ell} (1-2^{-2^j})^{-1}, \ \ \text{setting } \ell=k+1\\
&=& 1-\frac{1}{2}+2\sum_{k\ge 0} 2^{-2^k}  \prod_{0\le j<k} (1-2^{-2^j})^{-1}, \qquad \text{lumping the two sums into one}\\
&=& \frac12+\sum_{k\ge 0} 2^{-(2^0+\cdots+2^{k-1})} \prod_{0\le j<k} (1-2^{-2^j})^{-1}\\
&=& \frac12 + \sum_{k\ge 0} \prod_{0\le j\le k-1} \frac{1}{2^{2^j}-1},
\\
\end{eqnarray*}
which is the right-hand side.
