I was looking up the value of 52!, in preparation for a lecture, and I came across this expression:
$$52! = \int_{0}^{\infty} \frac{t^{52}}{e^t} dt $$
I checked it for some other values, and it appears that it's a general expression for the factorial of any integer. I.e. it appears that:
$$x! = \int_{0}^{\infty} \frac{t^{x}}{e^t} dt $$
How is this expression derived? I'm amazed by the connection between recursive multiplication of successively smaller integers, and an infinite integral involving all numbers raised to the power of the factorial.
Try doing integration by parts, considering $u = t^n$ and $dv/dt = e^{-t}$. In particular, we have $$ \int_0^\infty t^n e^{-t} dt = - \bigl[ t^n e^{-t} \bigr]_0^\infty + n \int_0^\infty t^{n-1} e^{-t} dt. $$ But the first term is $0$ for $n > 0$, since $0^n = 0$ and $t^n e^{-t} \to 0$ as $t \to \infty$.
Now write $I_n = \int_0^\infty t^n e^{-t} dt$. Can you do some sort of induction/iteration of the above argument to deduce that $I_n = n!$?