I ran across a few finite sums involving the gamma function while doing some research about special cases of the hypergeometric function.
They're all very similar, so I'll just post one of them.
$$\sum_{k=0}^{n} \frac{1}{k!(n-k)!\Gamma(k+\frac{1}{2}) \Gamma(a-k+ \frac{1}{2})} = \frac{\Gamma(a+n)}{n! \Gamma(n+ \frac{1}{2}) \Gamma(a)\Gamma(a+\frac{1}{2}) }$$
My first thought was just to write the sum as one big fraction and try to simplify. I didn't have much success doing that.
Perhaps the summand can be written in such a way that the terms of the series telescope.
I don't know if that qualifies for an answer, but if you multiply both sides by $\Gamma\big(a+\frac{1}{2}\big)\,\Gamma\big(n+\frac{1}{2}\big)$, you get this:
$$\sum_{k=0}^n\left(\begin{matrix}a-\frac{1}{2}\\k\end{matrix}\right)\left(\begin{matrix}n-\frac{1}{2}\\n-k\end{matrix}\right)=\left(\begin{matrix}a+n-1\\n\end{matrix}\right)$$
which is a special case of the Chu-Vandermonde identity.
Here, when $b$ is a nonnegative integer, $\left(\begin{matrix}a\\b\end{matrix}\right)$ is $\frac{\Gamma(a+1)}{b!\,\Gamma(a-b+1)}$.
To prove
$$\sum_{k=0}^n\left(\begin{matrix}x\\k\end{matrix}\right)\left(\begin{matrix}y\\n-k\end{matrix}\right)=\left(\begin{matrix}x+y\\n\end{matrix}\right)$$
for general complex $x$ and $y$, you start by fixing $x$ and considering this as a polynomial identity in $y$, so you only have to prove it for general complex $x$ and general integer $y\geq n$; then, fixing $y$, you only have to prove it for a general integer $x\geq n$.
It then follows from comparing the coefficients of $t^n$ in the expansions of $(1+t)^{x+y}$ and $(1+t)^x\,(1+t)^y$.