Why is $\int_0^\infty e^{-nx}x^{s-1}dx = \frac {\Gamma(s)}{n^s}$?

Question

Why is the following equation true?$$\int_0^\infty e^{-nx}x^{s-1}dx = \frac {\Gamma(s)}{n^s}$$ I know what the Gamma function is, but why does dividing by $n^s$ turn the $e^{-x}$ in the integrand into $e^{-nx}$? I tried writing out both sides in their integral forms but $n^{-s}$ and $e^{-x}$ don't mix into $e^{-nx}$. I tried using the function's property $\Gamma (s+1)=s\Gamma (s)$ but I still don't know how to turn it into the above equation. What properties do I need?

Answer 1

Note that the definition of the gamma function is: $$\Gamma(s)=\int_0^{\infty} x^{s-1}e^{-x}dx$$ So, taking $y=nx$ so that $dy=ndx$ and $x=\frac{y}{n}$ yields: $$\int_0^{\infty}e^{-nx}x^{s-1}=\int_0^{\infty}e^{-y}(\frac{y}{n})^{s-1}ndy=\frac{1}{n^s}\int_0^{\infty} y^{s-1}e^{-y}dy=\frac{\Gamma(s)}{n^s}$$

Answer 2

Hint: make the substitution $u = nx$ in the integral

$$\Gamma(s) = \int_0^\infty e^{-u}u^s \frac{du}{u}$$