Check whether $\sum\limits_{n=1}^{\infty}\frac{z^n}{(1-z^n)^k}=\sum\limits_{n=1}^{\infty}\sigma_{k-1}(n)z^n$ Is it true that 

$$\sum_{n=1}^{\infty}\frac{z^n}{(1-z^n)^k}=\sum_{n=1}^{\infty}\sigma_{k-1}(n)z^n$$

If yes, how can I prove it? 

$\displaystyle{%
\sum_{n = 1}^{\infty}{z^{n} \over \left(1 - z^{n}\right)^{k}}
=
\sum_{n = 1}^{\infty}\sigma_{\,k - 1}\left(n\right)z^{n}:\ {\large ?}
\quad\mbox{where}\quad
\sigma_{\,k - 1}\left(n\right)
\equiv
\sum_{d\,|\,n}d^{k - 1}}$

With $\left\vert z\right\vert < 1$:  

$$
{1 \over 1 - z} = \sum_{\ell = 0}^{\infty}z^{\ell}\,,
\quad
{1 \over \left(1 - z\right)^{2}} = \sum_{\ell = 1}^{\infty}\ell\,z^{\ell - 1}\,,
\quad
{1 \over \left(1 - z\right)^{3}} = {1 \over 2}
\sum_{\ell = 2}^{\infty}\ell\left(\ell - 1\right)z^{\ell - 2}
$$
$$
{1 \over \left(1 - z\right)^{4}} = {1 \over 3\cdot 2}
\sum_{\ell = 3}^{\infty}\ell\left(\ell - 1\right)\left(\ell - 2\right)
z^{\ell - 3}\,,
\quad
\ldots
$$
$$
\begin{array}{rcl}\hline\\
{1 \over \left(1 - z\right)^{k}}
& = &
{1 \over \left(k - 1\right)!}
\sum_{\ell = k - 1}^{\infty}\ell\left(\ell - 1\right)\ldots
\left(\ell - k + 2\right)z^{\ell - k + 1}
\\[3mm]& = &
{1 \over \left(k - 1\right)!}
\sum_{\ell = k - 1}^{\infty}{\ell! \over \left(\ell - k + 1\right)!}\,
z^{\ell - k + 1}
\\[3mm]& = &
\sum_{\ell = k - 1}^{\infty}{\ell \choose k - 1}z^{\ell - k + 1}
=
\sum_{\ell = 0}^{\infty}{\ell + k - 1\choose k - 1}z^{\ell}
\\ \\ \hline
\end{array}
$$

Then,

\begin{align}
\color{#ff0000}{\large\sum_{n = 1}^{\infty}{z^{n} \over \left(1 - z^{n}\right)^{k}}}
&=
\sum_{n = 1}^{\infty}z^{n}\sum_{\ell = 0}^{\infty}{\ell + k - 1\choose k - 1}
z^{n\ell}
=
\sum_{\ell = 0}^{\infty}{\ell + k - 1 \choose k - 1}
\sum_{n = 1}^{\infty}z^{n\ell + n}
\\[3mm]&=
\sum_{\ell = 0}^{\infty}{\ell + k - 1\choose k - 1}
{z^{\ell + 1} \over 1 - z^{\ell + 1}}
\\[3mm]&=
\sum_{\ell = 0}^{\infty}{\ell + k - 1 \choose k - 1}
z^{\ell + 1}\sum_{\ell' = 0}^{\infty}\left(z^{\ell + 1}\right)^{\ell'}
\sum_{n = 1}^{\infty}\delta_{n, \ell + \ell' + \ell\ell' + 1}
\\[3mm]&=
\sum_{n = 1}^{\infty}z^{n}
\sum_{\ell = 0}^{\infty}{\ell + k - 1 \choose k - 1}
\sum_{\ell' = 0}^{\infty}\delta_{n,\ell + \ell' + \ell\ell' + 1}
\\[3mm]&=
\sum_{n = 1}^{\infty}z^{n}
\sum_{\ell = 1}^{\infty}{\ell + k - 2 \choose k - 1}
\sum_{\ell' = 1}^{\infty}\delta_{n,\ell\ell'}
=
\color{#ff0000}{\large\sum_{n = 1}^{\infty}z^{n}
\sum_{\ell = 1 \atop \ell\,|\,n}^{n}{\ell + k - 2 \choose k - 1}}
\end{align}
