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? 

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 $k\geqslant1$, $$\gamma_1^k=1,\quad\gamma_2^k=1+k,\quad\gamma_3^k=1+\tfrac12k(k+1),\quad\gamma_4^k=1+k+\tfrac16k(k+1)(k+2).$$
