Complex series sum Show that for $|z|<1$, 
$$\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z}$$
and
$$\sum_{n=0}^\infty \frac{2^n z^{2^n}}{1+z^{2^n}}=\frac{z}{1-z}$$
As a hint is given to use the dyadic expansion of an integer.
I have no idea how to proceed.Please help.

The LHS of the first is
\begin{align}
&(z+z^3+z^5+z^7+\cdots)\\
&+(z^2+z^6+z^{10}+z^{14}+\cdots)\\
&+(z^4+z^{12}+z^{20}+z^{28}+\cdots)+\cdots.
\end{align}
The LHS of the second is
\begin{align}
&(z-z^2+z^3-z^4+z^5-\cdots)\\
&+(2z^2-2z^4+2z^6-2z^8+2z^{10}-\cdots)\\
&+(4z^4-4z^8+4z^{12}-4z^{16}+4z^{20}-\cdots)+\cdots.
\end{align}
Do these give a clue?
