Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ Show $\displaystyle\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ for $|z|<1$.
This is an additional problem for my complex analysis class and I've attempted it for a few hours but ended up taking wrong routes. All of my attempts I haven't used complex analysis at all and I don't see how I could here.
edit: this is meant for a BEGINNER complex analysis course so please try keep the solutions to that (if you could)
Any help would be great

Hint: Let $\displaystyle F(z)=\sum_{n\geq 0}\frac{z^{2^n}}{1-z^{2^{n+1}}}$. Put $\displaystyle G(z)=F(z)-\frac{z}{1-z}$. Compare $G(z)$ and $G(z^2)$.
