Sum of series $\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$ 
The problem is to find the sum of series $$\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$$
  to infinite terms, if $|x|<1$ 

I tried taking $\frac{x}{1-x^2}$ outside but could not solve it.
How to proceed further?

If $|x|<1$, for any $k\in\mathbb{N}$ we have
$$ \frac{x^{2^k}}{1-x^{2^{k+1}}} = x^{2^k}+x^{3\cdot 2^k}+x^{5\cdot 2^k}+\ldots\tag{A}$$
but given any $n\in\mathbb{N}^+$, there is a unique representation of $n$ as $2^k\cdot d$, with $k\in\mathbb{N}$ and $d\in 2\mathbb{N}+1$. It follows that:
$$ \sum_{k\geq 0}\frac{x^{2^k}}{1-x^{2^{k+1}}} = \sum_{n\geq 1} x^n = \color{red}{\frac{x}{1-x}}.\tag{B}$$
