I would like to know how one can prove this sum, based on a Brazilian math competition problem:
$$\sum_{i=1}^n\left(\prod_{q=0}^p (i+q)\right) = \frac{1}{p+2}\prod_{s=0}^{p+1}(n+s)$$
In order to try to solve it, I've managed to get:
If,
$$ S(x)=\sum_{i=0}^{n}x^{i+p}$$
Then,
$$\frac{\mathrm{d}^{p+1}S(x)}{\mathrm{d} x^{p+1}}=\sum_{i=0}^{n}\left(\left( \prod_{q=0}^{p}\left( i+ q \right)\right)x^{i-1}\right)$$
Yet I don't see how to proceed any further.
First, let prove by induction that for all $n \in \mathbb{N}^*$, and all $p \in \mathbb{N}$, $$\sum_{i=1}^n {p+i \choose i-1} = {n+p+1 \choose n-1} \quad \quad \quad (1)$$
For $n = 1$, it is ok. Now let's suppose that it is true for an integer $n \geq 1$ ; then you have, using the induction hypothesis and Pascal's formula, $$\sum_{i=1}^{n+1} {p+i \choose i-1} = \sum_{i=1}^n {p+i \choose i-1} + {n+p+1 \choose n} = {n+p+1 \choose n-1} + {n+p+1 \choose n} = {n+p+2 \choose n}$$
So the formula $(1)$ is true. Now you can prove your equality. Indeed
$$\sum_{i=1}^n \left( \prod_{q=0}^p (i+q)\right) = \sum_{i=1}^n \frac{(p+i)!}{(i-1)!} = (p+1)!\sum_{i=1}^n {p+i \choose i-1}$$
Using formula $(1)$, you get $$\sum_{i=1}^n \left( \prod_{q=0}^p (i+q)\right) = (p+1)!{n+p+1 \choose n-1} = \frac{1}{p+2}(p+2)! \frac{(n+p+1)!}{(n-1)!(p+2)!}$$ $$= \frac{1}{p+2} \times \frac{(n+p+1)!}{(n-1)!} = \frac{1}{p+2}\prod_{s=0}^{p+1}(n+s)$$