Showing $\frac{z}{1+z}+\frac{2z^2}{1+z^2}+...+\frac{2^{k}z^{2^k}}{1+z^{2k}}+...=\frac{z}{1-z}$ 
Prove that for $\left\lvert z \right\rvert<1$, $\dfrac{z}{1+z}+\dfrac{2z^2}{1+z^2}+...+\dfrac{2^{k}z^{2^k}}{1+z^{2k}}+...=\dfrac{z}{1-z}$. Also, justify any change in the order of summation.

This is a exercise from my textbook, but I have no idea. I have shown that: $$\dfrac{z}{1+z}+\dfrac{2z^2}{1+z^2}+...+\dfrac{2^{k}z^{2^k}}{1+z^{2k}}-\dfrac{z}{1-z}= \dfrac{2^{k+1}}{1-\dfrac{1}{z^{2^{k+1}}}}$$
but I do not know how to show $2^{k+1}$
 is growing 'slow enough' in order to make the whole fraction tending zero.
The hint from the book is: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+...+2^k$
Could you please give me some more hint? Thank you.

Note that for each $k \ge 1$,
$$\frac{2^k z^{2^k}}{1+z^{2^k}} = \frac{2^k z^{2^k}(1 - z^{2^k})}{(1 + z^{2^{k}})(1-z^{2^k})} = \frac{2^kz^{2^k}(1+z^{2^k} - 2z^{2^k})}{(1+z^{2^{k}})(1-z^{2^k})} = \frac{2^kz^{2^k}}{1-z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}. $$
Further, since $\lvert z \rvert < 1$, 
$$\lim_{k\to \infty} \frac{2^k z^{2^k}}{1 - z^{2^k}} = 0$$
With these two facts, you'll be able to prove the result.
