Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$

$$\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots$$

Add $\displaystyle \frac{-z}{1-z}$ to both sides. It's Telescoping series:

$$\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots=\\ =\frac{-z-z^2}{1-z^2}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots= \\ = \frac{-z^2}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots= \\ = \frac{-z^4-z^2}{1-z^4}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots=\frac{-z^4}{1-z^4}+\frac{z^4}{1-z^8}+ \cdots$$

$\textbf{Edit:}$ If you want to have finite sum rather than infinity. You can show that (like above - by cancelling terms) :

$$\frac{-z}{1-z}+\sum_{n=0}^{m}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z^{2^{m}}}{1-z^{2^{m}}}$$

Next calculate limit for $m \to \infty$ both sides.

$$\lim_{m \to \infty}\frac{-z}{1-z}+\sum_{n=0}^{m}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}$$

Finally (because $|z|<1$): $$\lim_{m \to \infty}\frac{-z^{2^{m}}}{1-z^{2^{m}}}=0$$

The first term is $z/(1-z^2)$ which is a series where every exponent is odd.
What are the exponents in the series of the second term?

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 $G(z^2)$.