Two different expansions of $\frac{z}{1-z}$ This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis.
Show that for $|z|<1$ one has $$\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}$$and 
$$\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots =\frac{z}{1-z}.$$
Justify any change in the order of summation. 
[Hint: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots +2^k$.]
I don't really know how to work this through. I know that $\frac{z}{1-z}=\sum_{n=1}^\infty z^n$ and each $n$ can be represented as a dyadic expansion, but I don't know how to progress from here. Any hints solutions or suggestions would be appreciated.

Hint and partial answer: Using this partition of integers and
$$\frac{z}{1-z}=\sum_{n=1}z^n=...$$
This series is absolute converging given $|z|<1$, thus changing the order of summation doesn't affect the final value. As a result:
$$...=\sum_{k=0}\left(\sum_{t\in A_k}z^t\right)=\sum_{t\in A_0}z^t+ \sum_{k=1}\left(\sum_{t\in A_k}z^t\right)=\sum_{s=0}z^{2s+1} + \sum_{k=1}\left(\sum_{s=0}z^{2^k(2s+1)}\right)=\\
z\sum_{s=0}z^{2s}+\sum_{k=1}z^{2^k}\left(\sum_{s=0}z^{2^k(2s)}\right)=\frac{z}{1-z^2}+\sum_{k=1}\frac{z^{2^k}}{1-z^{2^{k+1}}}$$
