How to prove this equality  $ t(1-t)^{-1}=\sum_{k\geq0} 2^k t^{2^k}(1+t^{2^k})^{-1}$? 
Prove the equality $\quad t(1-t)^{-1}=\sum_{k\geq0} 2^k t^{2^k}(1+t^{2^k})^{-1}$.

I have just tried to use the Taylor's expansion of the left to prove it.But I failed.
I don't know how the $k$ and $2^k$ in the right occur. And this homework appears after some place with the $Jacobi$ $Identity$ in the book $Advanced$ $Combinatorics$(Page 118,EX10 (2)).
Any hints about the proof ?Thank you in advance.

This is all for $|t|&lt1$.  Start with the geometric series
$$\frac{t}{1-t} = \sum_{n=1}^\infty t^n$$
On the right side, each term expands as a geometric series 
$$\frac{2^k t^{2^k}}{1+t^{2^k}} = \sum_{j=1}^\infty 2^k (-1)^{j-1} t^{j 2^k}$$
If we add this up over all nonnegative integers $k$, for each integer $n$ you get
a term in $t^n$ whenever $n$ is divisible by $2^k$, with coefficient $+2^k$ when
$n/2^k$ is odd and $-2^k$ when $n/2^k$ is even.  So if $2^p$ is the largest power of $2$ that divides $n$, the coefficient of $t^n$ will be $2^p -\sum_{k=0}^{p-1} 2^k = 1$. 
