Why does an improper integral turn into an answer with factorial?

Question

Suppose I have

$\int_{0}^{\infty}y^{2n+1}e^{-y}dy$

Why does this integral equal

$(2n+1)!$ ?

Could somebody please explain this?

Answer 1

Repeated integration by parts.

$\int_0^\infty y^{2n+1}e^{-y}dy = -y^{2n+1}e^{-y}|_0^\infty + (2n+1)\int_0^\infty y^{2n}e^{-y}dy = (2n+1)\int_0^\infty y^{2n}e^{-y}dy$.