Pretty Nice Integral

Consider,
$I= \displaystyle \int_{0}^{\infty} \dfrac{x^{n}\text{d}x}{e^{ax}-1} \\ I= \displaystyle \int_{0}^{\infty} \dfrac{x^{n}e^{-ax}dx}{1-e^{-ax}} \\ Now, \dfrac{e^{-ax}}{1-e^{-ax}} = e^{-ax} +e^{-2ax} +\cdots \infty \\ = \displaystyle \sum_{r=1}^{\infty}e^{-arx} \\ I= \displaystyle \sum_{r=1}^{\infty} \displaystyle \int_{0}^{\infty} x^{n}e^{-arx}dx \\ Set\quad arx=t; ar\text{d}x=dt \\ I=\displaystyle \sum_{r=1}^{\infty} \dfrac{1}{(ar)^{n+1}}\displaystyle \int_{0}^{\infty} t^{n}e^{-t}dt \\ \\ I=\dfrac{1}{a^{n+1}}\zeta(n+1)\Gamma(n+1) \\ answer: \boxed {\displaystyle \int_{0}^{\infty} \dfrac{x^{5}\text{d}x}{e^{5x}-1} = \dfrac{1}{5^{6}}\zeta(6)\Gamma(6)}$

We can use Laplace transforms also. Simple......:-)