Is there a combinatorial way to see the link between the beta and gamma functions? The Wikipedia page on the beta function gives a simple formula for it in terms of the gamma function.  Using that and the fact that $\Gamma(n+1)=n!$, I can prove the following formula:
$$
\begin{eqnarray*}
\frac{a!b!}{(a+b+1)!} & = & \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+1+b+1)}\\
 & = & B(a+1,b+1)\\
 & = & \int_{0}^{1}t^{a}(1-t)^{b}dt\\
 & = & \int_{0}^{1}t^{a}\sum_{i=0}^{b}\binom{b}{i}(-t)^{i}dt\\
 & = & \int_{0}^{1}\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}t^{a+i}dt\\
 & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\int_{0}^{1}t^{a+i}dt\\
 & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\left[\frac{t^{a+i+1}}{a+i+1}\right]_{t=0}^{1}\\
 & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\\
b! & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{a!(a+i+1)}
\end{eqnarray*}
$$
This last formula involves only natural numbers and operations familiar in combinatorics, and it feels very much as if there should be a combinatoric proof, but I've been trying for a while and can't see it. I can prove it in the case $a=0$:
$$
\begin{eqnarray*}
 &  & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(b+1)!}{0!(i+1)}\\
 & = & \sum_{i=0}^{b}(-1)^{i}\frac{b!(b+1)!}{i!(b-i)!(i+1)}\\
 & = & b!\sum_{i=0}^{b}(-1)^{i}\frac{(b+1)!}{(i+1)!(b-i)!}\\
 & = & b!\sum_{i=0}^{b}(-1)^{i}\binom{b+\text{1}}{i+1}\\
 & = & b!\left(1-\sum_{i=0}^{b+1}(-1)^{i}\binom{b+\text{1}}{i}\right)\\
 & = & b!
\end{eqnarray*}
$$
Can anyone see how to prove it for arbitrary $a$? Thanks!

Seven years later I found another way to attack this. Define $f(b, a) = \frac{a!b!}{(a+b+1)!}$ and $h(b, a) = \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}$. To connect the two, we define $g$ such that $g(0, a) = \frac{1}{a + 1}$ and $g(b + 1, a) = g(b, a) - g(b, a + 1)$ and prove by induction in $b$ that $f = g = h$. In each case the base case is straightforward and we consider only the inductive step.
$$
\begin{eqnarray*}
& & g(b + 1, a) \\
& = & g(b, a) - g(b, a + 1) \\
& = & f(b, a) - f(b, a + 1) \\
& = & \frac{a!b!}{(a+b+1)!} - \frac{(a+1)!b!}{(a+b+2)!} \\
& = & \frac{a!b!(a + b + 2)}{(a+b+2)!} - \frac{a!b!(a+1)}{(a+b+2)!} \\
& = & \frac{a!b!(b+1)}{(a+b+2)!} \\
& = & \frac{a!(b+1)!}{(a+b+2)!} \\
& = & f(b+1, a)\\
\end{eqnarray*}
$$
$$
\begin{eqnarray*}
& & g(b + 1, a) \\
& = & g(b, a) - g(b, a + 1) \\
& = & h(b, a) - h(b, a + 1) \\
& = & \left(\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\right) - \left(\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+2}\right) \\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\right) - \left(\sum_{i=0}^{b-1}\binom{b}{i}(-1)^{i}\frac{1}{a+i+2}\right) - (-1)^{b}\frac{1}{a+b+2}\\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\right) + \left(\sum_{i=1}^{b}\binom{b}{i-1}(-1)^{i}\frac{1}{a+i+1}\right) + (-1)^{b+1}\frac{1}{a+b+2}\\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\left(\binom{b}{i} + \binom{b}{i-1}\right)(-1)^{i}\frac{1}{a+i+1}\right) + (-1)^{b+1}\frac{1}{a+b+2}\\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\binom{b+1}{i}(-1)^{i}\frac{1}{a+i+1}\right) + (-1)^{b+1}\frac{1}{a+b+2}\\
& = & \sum_{i=0}^{b+1}\binom{b+1}{i}(-1)^{i}\frac{1}{a+i+1} \\
& = & h(b + 1, a) \\
\end{eqnarray*}
$$
