There are lots (an infinitude) of smooth functions that coincide with $f(n)=n!$ on the integers. Is there a simple reason why Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^{-t} dt$ is the "best"?  In particular, I'm looking for reasons that I can explain to first-year calculus students.
The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties 


$f(1)=1$;
$f(x+1)=xf(x)$ for every $x\geq 0$;
$\log f$ is a convex function. 


The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function. 



By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),\qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2\dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{m=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z, $$
($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=\gamma z$.

Edit: corrected index in the product.
