A truncated alternating sum of product of binomial terms While solving a question I came across the following alternating sum; $C(j,n): = \sum\limits_{i=j}^{n} (-1)^{i}\binom{n+1}{i+1} \binom{i}{j}$ where $j$ and $n$ are integers with $n \geq j \geq 0$. 
By hand I computed that $C(j, j+r) = (-1)^{j}$ for small positive integers $r$. I think that $C(j,n) = (-1)^j$ for any $n \geq j \geq 0$. But I couldn't prove it by induction or by using some other known identities. I would appreciate any suggestion or reference.  

\begin{align}
\sum_{i=j}^{n}(-1)^i\binom{n+1}{i+1}\binom{i}{j}&=(-1)^j\sum_{i=0}^{n}(-1)^i\binom{n+1}{i+1}[x^j](1-x)^i\\&=(-1)^j[x^j]\sum_{i=0}^{n}\binom{n+1}{i+1}(x-1)^i\\&=(-1)^j[x^j]\frac{x^{n+1}-1}{x-1}=(-1)^j.
\end{align}
