Splitting of integral with gamma function

Question

I was looking at the solution here and I do not understand this step: $$\sum_{n=0}^{\infty}\int_0^{\infty}x^me^{-x(n+1)}\;\mathrm{d}x =\Gamma(m+1)\sum_{n=0}^{\infty}\frac{1}{(n+1)^{m+1}}$$ I understand that $\sum_{n=0}^{\infty}\int_0^{\infty}x^me^{-x(n+1)}\;\mathrm{d}x=\sum_{n=0}^\infty \int_0^\infty x^me^{-x}e^{-xn}$, and that $\Gamma(m+1)=\int_0^\infty x^me^{-x}$, but I do not understand how the integral splits. Any hint would be greatly appreciated.

Answer 1

Substituting $u=x(n+1)$ gives $$ \int_0^{+\infty}x^me^{-(n+1)x}dx=\frac{1}{(n+1)^{m+1}}\int_0^{+\infty}u^me^{-u}du=\frac{\Gamma(m+1)}{(n+1)^{m+1}} $$

Answer 2

HINT:

In the integral $\int_0^\infty x^m e^{-(n+1)x}\,dx$ enforce the substitution $x =t/(n+1)$.

Can you finish now?