How does $\displaystyle \frac{e^{-x}}{1-e^{-x}}$ become $\sum_{k=1}^{\infty}e^{-kx}$? How does $\displaystyle \frac{e^{-x}}{1-e^{-x}}$ become $\sum_{k=1}^{\infty}e^{-kx}$?
I know $\displaystyle \frac{e^{-x}}{1-e^{-x}}=\frac{1}{e^x-1}=\left(\sum_{k=0}^{\infty} \frac{x^k}{k!}-1\right)^{-1}=\left(\sum_{k=1}^{\infty} \frac{x^k}{k!}\right)^{-1}$ and we know $\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!}=e^x$

You have if $|a|<1$:
$$
\sum_{i=0}^\infty a^i = \frac{1}{1-a}
$$
therefore:
$$
\frac{e^{-x}}{1-e^{-x}}= e^{-x}\sum_{i=0}^\infty \left({e^{-x}}\right)^i = \sum_{n=1}^\infty e^{-nx}.
$$
