Proving $\int_0^\infty e^{-ax}x^n\,dx = \frac{1}{a^{n+1}} \Gamma(n+1)$

Question

Prove that $$ \int_{0}^{\infty} \ e^ {-ax} x^{n} dx = \frac{1}{a^{n+1}} \Gamma(n+1) \qquad (n>-1, \, a>0). $$

My try:

Let $dv = e^{-ax}$ and $u = x^n$.

Then $v = -\frac{1}{a}e^{-ax}$ and $du = nx^{n-1}$.

So the integral is $ -\frac{1}{a}e^{-ax} x^n |_{\infty}^0 + \int_{\infty}^0 [ -\frac{1}{a}e^{-ax}\cdot nx^{n-1}]$. I'm stuck here.

Answer 1

Use substitution $y=ax$, so that the integral becomes $$\int_0^\infty e^{-y}\left(\frac{y}{a}\right)^n\ \left(\frac{dy}{a}\right)=\frac{1}{a^{n+1}}\int_0^\infty e^{-y}y^{(n+1)-1}\ dy=\frac{1}{a^{n+1}}\Gamma(n+1)$$