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.
 A: We have
$$
\eqalign{
  & \sum\limits_{j = 0}^n {\left( { - 1} \right)^{\,n - j} \left( \matrix{
  p + j \cr 
  j \cr}  \right)\left( \matrix{
  n + \beta  \cr 
  n - j \cr}  \right)}  =   \cr 
  &  = \sum\limits_{j = 0}^n {\left( { - 1} \right)^{\,n} \left( \matrix{
   - p - 1 \cr 
  j \cr}  \right)\left( \matrix{
  n + \beta  \cr 
  n - j \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{
   - p - 1 \cr 
  j \cr}  \right)\left( \matrix{
  n + \beta  \cr 
  n - j \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \left( \matrix{
   - p - 1 + n + \beta  \cr 
  n \cr}  \right) =   \cr 
  &  =  \left( \matrix{
  p - \beta  \cr 
  n \cr}  \right)\quad \left| \matrix{
  \;n \in Z \hfill \cr 
  \;\forall p,\beta  \hfill \cr}  \right. \cr} 
$$
where the steps are:
 - upper negation (always valid for integer $j$);
 - we can omit summming limits, because they are implicit in the two binomials;
 - convolution;
 - upper negation (always valid for integer $n$).   
A: 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
$$\binom xn=\left|\binom Xn\right|=\left|\binom Yn-\bigcup_{k=1}^m\mathcal A_k\right|=\left|\binom Yn\right|-\left|\bigcup_{k=1}^m\mathcal A_k\right|=\binom yn-\sum_{\emptyset\ne K\subseteq[m]}(-1)^{k-1}\left|\bigcap_{k\in K}\mathcal A_k\right|=\binom yn-\sum_{\emptyset\ne K\subseteq[m]}(-1)^{k-1}\binom{y-|K|}{n-k}=\binom yn-\sum_{k=1}^m(-1)^{k-1}\binom mk\binom{y-k}{n-k}=\sum_{k=0}^m(-1)^k\binom mk\binom{y-k}{n-k}=\sum_{k=0}^n(-1)^k\binom mk\binom{y-k}{n-k}=\sum_{k=0}^n(-1)^k\binom{y-x}k\binom{y-k}{n-k}.$$
A: Evaluating
$$\sum_{j=0}^n (-1)^{n-j} {p+j\choose j} {n+\beta\choose n-j}
= \sum_{j=0}^n (-1)^j {p+n-j\choose n-j} {n+\beta\choose j}$$
we write
$$\sum_{j=0}^n (-1)^j {n+\beta\choose j}
[z^{n-j}] (1+z)^{p+n-j}
\\ = [z^n] (1+z)^{p+n} \sum_{j=0}^n (-1)^j {n+\beta\choose j}
z^j (1+z)^{-j}.$$
Now we may  extend $j$ beyond $n$ because of the coefficient extractor
in front:
$$[z^n] (1+z)^{p+n} \sum_{j\ge 0} (-1)^j {n+\beta\choose j}
z^j (1+z)^{-j}
\\ = [z^n] (1+z)^{p+n}
\left(1-\frac{z}{1+z}\right)^{n+\beta}  
= [z^n] (1+z)^{p-\beta} = {p-\beta\choose n}.$$
