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.

Hint $\ $ Let $\rm\:N\to\infty\ $ in  $\rm\displaystyle\ \sum_{K\!\:=\!\:0}^{N-1}\!\ \frac{2^{\:\!K}\:\! t^{2^{\:\!K}}}{t^{2^K}\!+1} + \frac{t}{t-1}\: =\ \frac{2^{\:\!N}\:\! t^{2^{\:\!N}}}{t^{2^{\:\!N}}\!-1}\ =\: c\ t^{2^{\:\!N}} + \:\cdots\ $ by  telescopy.
Since $\rm\: t^{2^{\:\!N}} \to 0\:$ as $\rm\:N\to \infty,\:$ the desired formal power series equality follows.
See here for more on convergence of formal power series (beware many make errors here).
Remark $\ $ The telescopic proof is simply $\rm 2^{\:\!N}$ times below, for $\rm\:x = t^{2^N}$
$$\rm \frac{2\:x^2}{x^2-1} - \frac{x}{x-1}\: =\: \frac{x}{x+1}$$
