Vandermonde identity in a ring

Solution 1:

Here's one way to do the induction step for the equivalent reformulation: $$\begin{align*} (r+s-n)\prod_{k=0}^{n-1} (r+s-k) & =\sum_{p=0}^{n} \binom{n}{p} \left([r-p]+[s-(n-p)]\right)\cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n-p-1} (s-j) \\ & =\sum_{p=0}^{n} \binom{n}{p} \cdot \prod_{i=0}^{\color{red}{p}} (r-i) \cdot \prod_{j=0}^{n-p-1} (s-j) \\&\phantom{=}+ \sum_{p=0}^{n} \binom{n}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{\color{red}{n-p}} (s-j) \\ &=\binom{n}{n} \cdot \prod_{i=0}^{n} (r-i) + \sum_{p=\color{red}{1}}^{\color{red}{n}} \binom{n}{\color{red}{p-1}} \cdot \prod_{i=0}^{\color{red}{p-1}} (r-i) \cdot \prod_{j=0}^{\color{red}{n-p}} (s-j) \\&\phantom{=}+ \binom{n}{0}\prod_{j=0}^{n} (s-j) +\sum_{p=1}^{n} \binom{n}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n-p} (s-j) \\ &=\binom{n+1}{n+1} \cdot \prod_{i=0}^{n} (r-i) \\ &\phantom{=}+ \sum_{p=1}^{n} \left[\binom{n}{p-1}+\binom{n}{p}\right] \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n\color{red}{+1}-p\color{red}{-1}} (s-j) \\&\phantom{=}+ \binom{n+1}{0}\prod_{j=0}^{n} (s-j) \\ &= \sum_{p=0}^{n+1} \binom{n+1}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n+1-p-1} (s-j). \end{align*}$$