How to simplify this combinatorial sum

Let $x$ is real number and $n$ is integer. How to simplify this sum $\sum_k(-1)^k\binom{x+k}{k}\binom{x+1}{n-k}$? I tried different integers $x$ and get zero, but I don't know the result for arbitrary real $x$

\begin{align} \sum_k(-1)^k\binom{x+k}{k}\binom{x+1}{n-k} \stackrel1= \sum_k\binom{-x-1}{k}\binom{x+1}{n-k} \stackrel2= \binom0n =\begin{cases}0 & n > 0\\1 & n = 0\end{cases} \end{align}

1. Here, we apply the identity $\binom{m}j=(-1)^j\binom{-(m-j+1)}{j}$, where $m\gets x+k$ and $j\gets k$.

2. This is the Chu-Vandermonde identity, $\sum_k \binom{a}k\binom{b}{n-k}=\binom{a+b}n$, valid for all $a,b\in \mathbb C$.