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Another derivation of Euler's integral formula
$n! = \int_0^\infty x^n e^{-x} \, dx$
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\\
Inna Zakharevich (reported by Noam D. Elkies)
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Euler's integral formula $\int_0^\infty x^n e^{-x} \, dx$ for $n!$
is usually proved by repeated integration by parts.
Inna Zakharevich, a Math~55a student, didn't know the formula
and took a few minutes to find the following alternative derivation.
Begin with the familiar definite integral
$$
\int_0^\infty e^{-ax} \, dx
= \left[ - \frac{e^{-ax}}{a} \right]_{x=0}^\infty
= \frac1a \; ,
$$
valid for all $a>0$.
Differentiate $n$ times with respect to~$a$ to obtain
$$
\int_0^\infty (-x)^n e^{-ax} \, dx
= (-1)^n \frac{n!}{a^{n+1}} \; .
$$
Now simply take $a=1$!
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