TITLE: Intuition for the definition of the Gamma function?
QUESTION [304 upvotes]: In these notes by Terence Tao is a proof of Stirling's formula.  I really like most of it, but at a crucial step he uses the integral identity
$$n! = \int_{0}^{\infty} t^n e^{-t}  dt$$
, coming from the Gamma function.  I have a mathematical confession to make: I have never "grokked" this identity.  Why should I expect the integral on the right to give me the number of elements in the symmetric group on $n$ letters?  
(It's not that I don't know how to prove it.  It's quite fun to prove; my favorite proof observes that it is equivalent to the integral identity $\int_{0}^{\infty} e^{(x-1)t}  dt = \frac{1}{1 - x}$.  But if someone were to ask me, "Yes, but why, really?" I would have no idea what to say.)
So what are more intuitive ways of thinking about this identity?  Is there a probabilistic interpretation?  What kind of random variable has probability density function $\frac{t^n}{n!} e^{-t}$?  (What does this all have to do with Tate's thesis?)
As a rough measure of what I'm looking for, your answer should make it obvious that $t^n e^{-t}$ attains its maximum at $t = n$.
Edit:  The kind of explanation I'm looking for, as I described in the comments, is similar to this explanation of the beta integral.

REPLY [15 votes]: Geometric approach
Note that $\frac{t^n}{n!}$ is the volume of the set $S_t=\{(t_1,t_2,\dots,t_n)\in\mathbb R^{n}\mid t_i\geq 0\text{ and } t_1+t_2+\cdots+t_n\leq t\}$.
So we can perform a change of variables in the integral, replacing $t=t_1+\dots + t_{n+1}$, as:
$$\begin{align}\int_{0}^\infty \frac{t^n}{n!}e^{-t}\,dt &= \int_{0}^{\infty}\left(\int_{(t_i)\in S_t}\,dt_1\dots dt_n\right)e^{-t}\,dt \\&= \int_{t_1,t_2,\dots,t_n,t_{n+1}} e^{-(t_1+\dots +t_{n+1})}\,dt_1\dots dt_{n+1}
\end{align}$$
Where all variables go from $0$ to $\infty$. But this can clearly be factored as the $$\left(\int_{0}^\infty e^{-t}\,dt\right)^{n+1}=1$$
So, what's happening is really in $n+1$ dimensions - the linear map $\mathbb R^{n+1}\to\mathbb R^{n+1}$ defined as $$(t_1,\dots,t_n,t_{n+1})\mapsto\left(t_1,\dots,t_n,\sum_{i=1}^{n+1}t_i\right)$$
preserves volumes.
And then using the property $e^{x+y}=e^xe^y$.

Moment generating function approach
The exponential random variable $T$ with $P(T<t)=1-e^{-t}$ for $t\geq 0$ has:
$$E(T^n)=\int_{0}^{\infty} t^ne^{-t}\,dt$$
So we get the moment generating function: $$M_T(x) = \sum \frac{E(T^n)}{n!}x^n$$
But we also have:
$$M_T(x) = E(e^{Tx}) = \int_0^{\infty} e^{tx}e^{-t}\,dt = \int_0^{\infty} e^{-t(1-x)}\,dt = \frac{1}{1-x}$$
This means that $E(T^n)=n!$, which is the result you want.
I suppose this could be written without reference to probability as: If $I_n$ is your integral, then, since everything is positive and convergent, then for $0<x<1$ you can show:
$$\sum \frac{I_n}{n!}x^n = \int_0^{\infty} \left(\sum \frac{(xt)^n}{n!}\right)e^{-t}\,dt = \int_0^{\infty} e^{-t(1-x)}\,dt = \frac{1}{1-x}$$