How prove this $\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}=\sum_{n=1}^{\infty}\frac{q^n}{(1-q^n)^2}$ show that

$$\sum_{n=1}^{\infty}\dfrac{nq^n}{1-q^n}=\sum_{n=1}^{\infty}\dfrac{q^n}{(1-q^n)^2}$$
  where $|q|<1$

it seem use

$$\dfrac{q^n}{1-q^n}=\sum_{k=0}^{\infty}q^{n(k+1)}$$ But at last it can't work

maybe this problem have more nice methods.Thank you 
 A: \begin{align}
\sum_{m = 0}^{\infty}q^{nm}
&= {1 \over 1 - q^{n}}
\\[1cm]&\mbox{Derive both members respect}\ q
\\
\sum_{m = 0}^{\infty}nm\,q^{nm - 1}
&=
-\,{1 \over \left(1 - q^{n}\right)^{2}}\,\left(-nq^{n - 1}\right)
\\
\sum_{m = 1}^{\infty}m\,q^{nm}
&=
{q^{n} \over \left(1 - q^{n}\right)^{2}}
\\[1cm]&\mbox{Sum in both members for}\ n \geq 1
\\
\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}m\,q^{nm}
&=
\sum_{n = 1}^{\infty}{q^{n} \over \left(1 - q^{n}\right)^{2}}
\\
\sum_{m = 1}^{\infty}m\sum_{n = 1}^{\infty}\,q^{nm}
&=
\sum_{n = 1}^{\infty}{q^{n} \over \left(1 - q^{n}\right)^{2}}
\\
\sum_{m = 1}^{\infty}m\,{q^{m} \over 1 - q^{m}}
&=
\sum_{n = 1}^{\infty}{q^{n} \over \left(1 - q^{n}\right)^{2}}
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\sum_{n = 1}^{\infty}{nq^{n} \over 1 - q^{n}}
\color{#000000}{\ =\ }
\sum_{n = 1}^{\infty}{q^{n} \over \left(1 - q^{n}\right)^{2}}
\quad}
\\ \\ \hline
\end{array}
$$
A: \begin{align*}
\sum_{n=0}^\infty \frac{n x^n}{1-x^n } &= \sum_{n=0}^\infty -x \frac{d}{dx} \log(1-x^n)\\ 
 &= x \frac{d}{dx} \left(\sum_{n=0}^\infty \sum_{k=1}^\infty \frac{x^{n k }}{k } \right ) \\ 
 &= x \frac{d}{dx} \left(\sum_{k=1}^\infty \frac{1}{k(1-x^k )} \right ) \\ 
 &= \sum_{k=1}^\infty \frac{x^k}{(1-x^k)^2}\\ 
\end{align*}
