For $n \in \mathbb{N}$, prove that $\int_{0}^{\infty}x^{n}e^{-x}dx=n!$

Question

Well I am aware of the exponential representation of a factorial using summation but does this hold true for an integral as well? How should I prove this? Is it simple induction or is there more to this?

Answer 1

Let $I_n = \int_0^\infty x^ne^{-x}dx$. One has

$$I_1 = \int_0^\infty e^{-x}dx = 1.$$

$$I_n = -x^ne^{-x} \big\vert_0^\infty + n\int_0^\infty e^{-x}x^{n-1}dx = nI_{n-1}.$$