TITLE: Prove the series $\sum_{n=1}^{\infty}{x^{n-1}\over(n-1)!}\cdot{e^{-xn}-1\over e^{xn}-1}=-e^{-x(1-e^{-x})}$
QUESTION [7 upvotes]: Show that,

$$\sum_{n=1}^{\infty}{x^{n-1}\over(n-1)!}\cdot{e^{-xn}-1\over e^{xn}-1}=-e^{-x(1-e^{-x})}$$

My try:
We know 
$$\sum_{n=1}^{\infty}{x^{n-1}\over (n-1)!}=e^x$$
$$\sum_{n=1}^{\infty}{e^{-xn}-1\over e^{xn}-1}={1\over 1-e^{x}}$$
How do I use these two formulae to arrive  at the top formula?
Let and try letting $x=\ln{y}$ then we have
$$\sum_{n=1}^{\infty}{\ln{y}^{n-1}\over (n-1)!}\cdot{y^{-n}-1\over y^n-1}=-y^{{1\over y}-1}$$
Still can't see anything obvious step to do next, any hints?

REPLY [7 votes]: Note that $$\frac{e^{-xn}-1}{e^{xn}-1}=\frac{1}{e^{nx}}\frac{e^{-xn}-1}{1-e^{-nx}}=-e^{-nx}
 $$ hence $$\sum_{n\geq1}\frac{x^{n-1}}{\left(n-1\right)!}\frac{e^{-xn}-1}{e^{xn}-1}=-\sum_{n\geq1}\frac{x^{n-1}}{\left(n-1\right)!}e^{-xn}
 $$ $$=-e^{-x}\sum_{n\geq0}\frac{x^{n}}{n!}e^{-xn}=\color{red}{-\exp\left(-x+xe^{-x}\right)}.$$