How do I prove this complex-looking sum-product equation? The identity/theorem/equation:
$$\sum_{i=1}^{n} {\prod_{j=0}^k{(i+j)}} = \frac{\prod_{i=0}^{k+1}{(n+i)}}{k+2}$$
I know it has to with brute-forcedly expanding one of the sides, but how? Is there a simpler way to do it?

Clearly all numbers satisfies: $$\prod\limits_{j=0}^k (i+j) = {i+k \choose k+1} \cdot (k+1)!$$
Then $$\sum\limits_{i=1}^n \prod\limits_{j=0}^k (i+j) = (k+1)! \sum\limits_{i=1}^n {i+k \choose k+1}$$
By other hand $$\frac{\prod\limits_{i=0}^{k+1} (n+i)}{k+2}={ n+k+1 \choose k+2} \cdot (k+1)!$$
Then we have $${n+k+1 \choose k+2}=\sum \limits_{i=1}^n {i+k \choose k+1}$$
Christmas Stocking Theorem.
