A finite summation of double binomial coefficients I find the following identity and have checked on Mathematica, while I have no idea how to prove it:
$$\sum_{j=0}^n(-1)^{n-j}\binom{p+j}{j}\binom{n+\beta}{n-j}=\binom{p-\beta}{n}, \quad \beta>-1, \quad p>\beta-1.$$
It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either.

First, if $n,x,y$ are integers, $0\le n\le x\le y$, then
$$\binom xn=\sum_{k=0}^n(-1)^k\binom{y-x}k\binom{y-k}{n-k}.\tag1$$
This is just the inclusion-exclusion principle. If $X,Y$ are sets, $X\subseteq Y$, $|X|=x$, $|Y|=y$, then $\binom xn$ is the number of $n$-element subsets of $X$, which we think of as being sifted out of the collection of all $n$-element subsets of $Y$, and $\binom{y-x}k$ is the number of $k$-element subsets of $Y\setminus X$, and $\binom{y-k}{n-k}$ is the number of $n$-element subsets of $Y$ containing a given $k$-element subset.
Of course, for a fixed integer $n\ge0$, since $(1)$ is a polynomial identity in $x$ and 
$y$ which holds for all integers $y\ge x\ge n$, it also holds for all real or complex values of $x$ and $y$.
Setting $x=p-\beta$ and $y=p+n$ in $(1)$, we get
$$\binom{p-\beta}n=\sum_{k=0}^n(-1)^k\binom{n+\beta}k\binom{p+n-k}{n-k}.\tag2$$
Finally, setting $k=n-j$ in $(2)$, we get your identity:
$$\binom{p-\beta}n=\sum_{j=0}^n(-1)^{n-j}\binom{n+\beta}{n-j}\binom{p+j}j$$
for natural $n$ and arbitrary $\beta$ and $p$.
P.S. Here is a detailed explanation of $(1)$. If $S$ is a set, $\binom Sn$ is the set of all $n$-element subsets of $S.$ Suppose $n,x,y$ are integers, $0\le n\le x\le y$, and let $X,Y$ be sets, $|X|=x$, $|Y|=y$, $X\subseteq Y$. Let $m=y-x$, $Y\setminus X=\{y_1,\dots,y_m\}$, $[m]=\{1,\dots,m\}$. For $k\in[m]$ let $\mathcal A_k=\{A\in\binom Yn:y_k\in A\}$. Then
$$\binom xn=\left|\binom Xn\right|=\left|\binom Yn-\bigcup_{k=1}^m\mathcal A_k\right|=\left|\binom Yn\right|-\left|\bigcup_{k=1}^m\mathcal A_k\right|=\binom yn-\sum_{\emptyset\ne K\subseteq[m]}(-1)^{k-1}\left|\bigcap_{k\in K}\mathcal A_k\right|=\binom yn-\sum_{\emptyset\ne K\subseteq[m]}(-1)^{k-1}\binom{y-|K|}{n-k}=\binom yn-\sum_{k=1}^m(-1)^{k-1}\binom mk\binom{y-k}{n-k}=\sum_{k=0}^m(-1)^k\binom mk\binom{y-k}{n-k}=\sum_{k=0}^n(-1)^k\binom mk\binom{y-k}{n-k}=\sum_{k=0}^n(-1)^k\binom{y-x}k\binom{y-k}{n-k}.$$
