A truncated alternating sum of product of binomial terms

Here we have Chu-Vandermonde's Identity in disguise.

We obtain \begin{align*} \color{blue}{\sum_{i=j}^n}&\color{blue}{(-1)^i\binom{n+1}{i+1}\binom{i}{j}}\\ &=\sum_{i=0}^{n-j}(-1)^{i+j}\binom{n+1}{i+j+1}\binom{i+j}{i}\tag{1}\\ &=(-1)^j\sum_{i=0}^{n-j}\binom{n+1}{n-j-i}\binom{-j-1}{i}\tag{2}\\ &=(-1)^j\binom{n-j}{n-j}\tag{3}\\ &\,\,\color{blue}{=(-1)^j} \end{align*}

and the claim follows.

Comment:

• In (1) we shift the index to start with $i=0$ and we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

• In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (3) we apply Chu-Vandermonde's identity.