How to find the sum of $i(i+1)\cdots(i+k)$ for fixed $k$ between $i = 1$ and $n$? I learned that $$\sum \limits_{i=1}^n i(i+1) = \frac{n(n+1)(n+2)}{3}$$ or in general $$\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \frac{n(n+1)\dots (n+k+1)}{k+2}$$ From a mathematical standpoint why is this true? I'm not asking for inductive proof. I am asking if you only given the left hand side, how would you go about writing a closed form expression for the sum?

$$
S=\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!}
$$
$$
\frac{S}{(k+1)!}=\sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!(k+1)!}=\sum_{i=0}^{n-1}\binom{i+k+1}{i}
$$
$$
\frac{S}{(k+1)!}=\binom{k+1}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\
=\binom{k+2}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\
=\binom{k+3}{1} + \binom{k+3}{2} + \dots + \binom{n+k}{n-1} \\
=\binom{n+k+1}{n-1}
$$
the above uses $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$
$$
S=\binom{n+k+1}{n-1}*(k+1)!=\frac{(n+k+1)!(k+1)!}{(n-1)!(k+2)!}= \frac{n(n+1)\dots (n+k+1)}{k+2}
$$
