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## Homework Statement

Prove [tex]\int_{0}^{\infty}x^{n}e^{-x} dx = n![/tex]

## Homework Equations

[tex]0! = 1[/tex] (by convention)

## The Attempt at a Solution

Basic step:

[tex]

n=0 \\

\int_{0}^{\infty}x^{0}e^{-x} dx\ = 0! = 1\\

\int_{0}^{\infty}e^{-x} dx\ = -[e^{-\infty}-e^{0}]\\

-[e^{-\infty}-e^{0}] = -[\frac{1}{e^{\infty}}-1]\\

-[\frac{1}{e^{\infty}}-1] = -[0-1] = 1[/tex]

Therefore the statement is true for n = 0.

Assuming that [tex]\int_{0}^{\infty}x^{n}e^{-x} dx = n![/tex] is true,

I need to show [tex]\int_{0}^{\infty}x^{n+1}e^{-x} dx = (n+1)![/tex]

I don't really know what to do from here. I emailed the TA and he said to integrate by parts, but we haven't been taught that , and he hasn't replied again so if someone could help me out that would be great! Thanks!