showing $\frac{x}{e^{x}-1} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $ I want to show
$\frac{x}{e^{x}-1} = -\sum_{k=0}^{\infty} x e^{kx} = \frac{\ln(e^x)}{e^x-1} = -\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $
The first step seems trivial, what i am confused is the process of 
\begin{align}
-\sum_{k=0}^{\infty}xe^{kx} = \frac{\ln(e^x)}{e^x-1}
\end{align}
and the last step 
\begin{align}
-\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n}
\end{align}
Can you give me some idea for these steps? 
 A: 
In order to show the identity
  \begin{align*}
\frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{\left(1-e^x\right)^{n-1}}{n}\qquad\qquad |1-e^x|<1\tag{1}
\end{align*}

we recall that
\begin{align*}
  \ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\qquad\qquad\qquad |x|<1
  \end{align*}

We obtain
  \begin{align*}
x&=\ln(e^x)=\ln(1-(1-e^x))\\
&=-\sum_{n=1}^\infty \frac{1}{n} (1-e^x)^n\qquad\qquad\qquad\qquad\qquad |1-e^x|<1\\
\end{align*}
  and after division by $1-e^x$ with $x \neq 0$ we get (1)
  \begin{align*}
  \frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{1}{n} (1-e^x)^{n-1}\qquad\qquad\qquad\qquad x<\ln 2,x\neq 0
  \end{align*}

Note, it is not necessary to consider  a geometric series expansion. In fact the representation in this case is different from the series representation in (1) since the radius of convergence is different.
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=0}^\infty xe^{kx}\qquad\qquad |e^x|<1\quad\text{resp.}\quad x<0
\end{align*}
A: As $x=\ln({e^x})$ ,
$$
\begin{align}
-\sum_{k=0}^{\infty}xe^{kx}\end{align} =\frac{x}{e^x-1}= \frac{\ln(e^x)}{e^x-1}
$$
And as $\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ ,
$$
-\frac{\ln[1-(1-e^x)]}{1-e^x} = -\frac{(-\sum_{n=1}^{\infty} \frac{(1-e^x)^n}{n})}{1-e^x}=\frac{1}{1-e^x}.\sum_{n=1}^{\infty} \frac{(1-e^x)^n}{n}=\sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n}
$$
