Proving Integral with Riemann Zeta Function

Question

I need some help showing that

$$\int_0^{\infty}\frac{x^se^xdx}{e^x-1} = s!\zeta(s).$$

I tried doing a $u$ sub where $u=e^x$, and then I tried integrating by parts, but I was left with the integral $\int_1^{\infty}\frac{s(\ln u)^{s-1}du}{u(u-1)}$, and I'm fairly certain this isn't the way to go. I know that $\Gamma(n) = \int_0^{\infty}{x^{n-1}}{e^xdx}$, but I don't know how I can apply this identity to this proof, if at all.

Answer 1

Hint: Employ the geometric series and the evaluation $$ \frac{{s!}}{{n^{s+1} }} = \int_0^{ + \infty } {e^{ - nx} x^{s} dx} . $$ (Can you prove this integral representation?)