How to show that for a complex number $s$ with $Re (s) > 1$, one has $$\sum_{k=1}^{\infty}\frac1{k^s} = \frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$
where $\Gamma(z) = \int_0^{+\infty} t^{z-1}\mathrm{e}^{-t}\mathrm{d}t$
For info this is the Riemann zeta function
Your claim is off. The correct result is:
\begin{align*} \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1}\, \mathrm{d}x &= \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{\infty} e^{-nx} \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-nx} x^{s-1}\, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s}\\ &= \Gamma(s) \sum_{n=1}^{\infty} \frac{1}{n^s}\\ &= \Gamma(s) \zeta(s) \end{align*}
As a side note,
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1}\, \mathrm{d}x = \Gamma(s) \eta(s)$$
where $\eta$ is the Dirichlet's eta function.