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? 
 A: One can start from the identity
$$
\sum_{n=1}^{\infty}\frac{z^n}{(1-z^n)^k}=\sum_{n\geqslant1}z^n\sum_{i_1\geqslant0}z^{ni_1}\cdots\sum_{i_k\geqslant0}z^{ni_k}=\sum_{n,i_1,\ldots,i_k}z^{n(1+i_1+\cdots+i_k)}=\sum_{N\geqslant1}\gamma_N^kz^N,
$$
where $\gamma_N^k$ is the size of the set $$\{(n,i_1,\ldots,i_k)\mid\,(1+i_1+\cdots+i_k)\cdot n=N,\,n\geqslant1,\,\forall1\leqslant j\leqslant k,\, i_j\geqslant0\}.$$
Thus, $n$ divides $N$ hence $d=N/n$ divides $N$. For every $d$, the number of $k$-tuples $(i_1,\ldots,i_k)$ such that $i_1+\cdots+i_k=d-1$ and every $i_j$ is nonnegative is the coefficient of $x^{d-1}$ in the series
$$
\left(\sum_{i\geqslant0}x^i\right)^k=\frac1{(1-x)^k}=\sum_{m\geqslant0}{m+k-1\choose k-1}x^m.
$$
This shows that, for every $N\geqslant1$,
$$
\gamma_N^k=\sum_{d\mid N}{d+k-2\choose k-1}.
$$
Thus, $(\gamma^k_N)_N$ is not the sequence $(\sigma_{k-1}(N))_N$ except if $k=1$ or $k=2$, for example, if $k=3$,
$$
\gamma_N^3=\sum_{d\mid N}\tfrac12d(d+1)=\tfrac12\sigma_2(N)+\tfrac12\sigma_1(N).
$$
More generally, $\gamma_N^k$ is a barycenter of the coefficients $\sigma_{1}(N)$, $\sigma_{2}(N)$, ..., $\sigma_{k-1}(N)$.
Finally, note that, for every $k\geqslant1$, $$\gamma_1^k=1,\quad\gamma_2^k=1+k,\quad\gamma_3^k=1+\tfrac12k(k+1),\quad\gamma_4^k=1+k+\tfrac16k(k+1)(k+2).$$
A: $\displaystyle{%
\sum_{n = 1}^{\infty}{z^{n} \over \left(1 - z^{n}\right)^{k}}
=
\sum_{n = 1}^{\infty}\sigma_{\,k - 1}\left(n\right)z^{n}:\ {\large ?}
\quad\mbox{where}\quad
\sigma_{\,k - 1}\left(n\right)
\equiv
\sum_{d\,|\,n}d^{k - 1}}$

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}
A: This identity is true only for $k=1,2$ and for $|z|<1$. You can use this theorem. 

Sum by Curves Theorem. An absolutely convergent series $\sum a_{mn}$ itself converges. Moreover, If $S_1\subseteq S_2\subseteq S_3\subseteq ...\subseteq\mathbb{N}\times\mathbb{N}$ is a nondecreasing sequence of finite sets having the property that for any $m,n$ there is a $\ell$ such that $$\mathbb{N}_m\times\mathbb{N}_n\subseteq S_\ell\subseteq S_{\ell+1}\subseteq S_{\ell+2}\subseteq ...,$$ then the sequence $s_\ell$ convergence, where $s_\ell$ is the finite sum $s_\ell:=\sum_{(m,n)\in S_\ell}a_{mn},$ and furthermore $\sum a_{mn}=\lim s_\ell.$

Let $S_\ell=T_1\bigcup...\bigcup T_\ell,  T_\ell=\lbrace(m,n)\in\mathbb{N}\times\mathbb{N};mn=\ell\rbrace$
for $k=1$ note that $$\frac{1}{1-z^n}=\sum_{m=0}^{\infty}z^{mn}=\sum_{m=1}^{\infty}z^{n(m-1)}$$
$$\sum_{n=1}^{\infty}\frac{z^n}{1-z^n}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}z^{mn}=\sum_{n=1}^{\infty}\sigma_0(n)z^n.$$
for $k=2$ note that $$\frac{1}{(1-z^n)^2}=\sum_{m=1}^{\infty}mz^{n(m-1)}$$
$$\sum_{n=1}^{\infty}\frac{z^n}{(1-z^n)^2}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}mz^{mn}=\sum_{n=1}^{\infty}\sigma_1(n)z^n.$$
A: What you can prove is that
$$
\sum_{n=1}^\infty \sigma_k(n)z^n = \sum_{m=1}^\infty f_k(z^m)
$$
where $f_k(u) = \left(u\dfrac{d}{du}\right)^k \dfrac{u}{1-u}$.
For $k\in\{0,1\}$ this is your formula, but not anymore for $k \geq 2$.
Proof.
$$
\sum_{n=1}^\infty \sigma_k(n)z^n = \sum_{n=1}^\infty \sum_{r \mid n}r^k z^n = \sum_{r=1}^\infty \sum_{m=1}^\infty r^k z^{rm} = \left(z\frac{d}{dz}\right)^k\sum_{r=1}^\infty \sum_{m=1}^\infty \frac{z^{rm}}{m^k}
$$
and
$$
\sum_{r=1}^\infty \sum_{m=1}^\infty \frac{z^{rm}}{m^k} = \sum_{m=1}^\infty \frac{1}{m^k}\frac{z^m}{1-z^m}.
$$
