For this exercise, I need to prove that the following sum
$\sum\limits_{i=0}^{n} \dfrac{(-1)^{n-i}(k+i)!}{(n-i)!(i!)^2}$
is equal to zero for $k = 0,1,...,n-1$ and one when $k=n$.
I already tried to solve this for some small values of $n$ to get an understanding of how I can prove this but I have absolutely no clue.
Starting from
$$\sum_{q=0}^n (-1)^{n-q} \frac{(k+q)!}{(n-q)! \times q!^2}$$
we find
$$k!\sum_{q=0}^n (-1)^{n-q} {k+q\choose q} \frac{1}{(n-q)! \times q!} \\ = \frac{k!}{n!} \sum_{q=0}^n (-1)^{n-q} {k+q\choose q} {n\choose q} \\ = \frac{k!}{n!} \sum_{q=0}^n (-1)^{n-q} {k+q\choose q} [z^{n-q}] (1+z)^n \\ = \frac{k!}{n!} [z^n] (1+z)^n \sum_{q=0}^n (-1)^{n-q} {k+q\choose q} z^q \\ = \frac{k!}{n!} (-1)^n [z^n] (1+z)^n \sum_{q\ge 0} (-1)^{q} {k+q\choose q} z^q \\ = \frac{k!}{n!} (-1)^n [z^n] (1+z)^n \frac{1}{(1+z)^{k+1}} \\ = \frac{k!}{n!} (-1)^n [z^n] (1+z)^{n-k-1} .$$
Now for $0\le k\le n-1$ this is clearly zero and for $k=n$ it evaluates to
$$(-1)^n [z^n] \frac{1}{1+z} = (-1)^{2n} = 1.$$