Proof of formula $\sum_{i=1}^n {}_{i+k}P_{k+1} = \sum_{i=1}^n \prod_{j=0}^k (i+j) = \frac{(n+k+1)!}{(n-1)!(k+2)}$

Question

I want to learn proof of

$$\sum_{i=1}^n {}_{i+k}P_{k+1} = \sum_{i=1}^n \prod_{j=0}^k (i+j) = \frac{(n+k+1)!}{(n-1)!(k+2)}$$

Can there be found an elementary algebraic proof that does not require high level of math?

I wrote $$\sum_{i=1}^n \prod_{j=0}^k (i+j)=\sum_{i=1}^n \left(i(i+1)(i+2)(i+3)\cdots (i+k) \right)=\sum_{i=1}^n \frac {(i+k)!}{(i-1)!}$$

Did I start right?

Answer 1

Yeah, great start, now divide both sides by $(k+1)!$ and use the hockey stick identity \begin{eqnarray*} \sum_{i=1}^{n} \binom{i+k}{k+1} =\binom{n+k+1}{k+2}. \end{eqnarray*}