Prove that $\sum\limits_{i=1}^{n-k} (-1)^{i+1} \cdot \frac{(k-1+i)!}{k! \cdot(i-1)!} \cdot \frac{n!}{(k+i)! \cdot (n-k-i)!}=1$ 
Prove that, for $n>k$, $$\sum\limits_{i=1}^{n-k} (-1)^{i+1} \cdot \frac{(k-1+i)!}{k! \cdot(i-1)!} \cdot \frac{n!}{(k+i)! \cdot (n-k-i)!}=1$$

I found this problem in a book at the library of my my university, but sadly the book doesn't show a solution. I worked on it quite a long time, but now I have to admit that this is above my (current) capabilities. Would be great if someone could help out here and write up a solution for me.

Note first that two factorials nearly cancel out, leaving us with $\frac1{k+i}=\int_0^1t^{k+i-1}\mathrm dt$, hence the sum to be computed is
$$
S=\sum_{i=1}^{n-k}(-1)^{i+1}n{n-1\choose k}{n-k-1\choose i-1}\int_0^1t^{k+i-1}\mathrm dt
$$
Second, note that
$$
\sum_{i=1}^{n-k}(-1)^{i+1}{n-k-1\choose i-1}t^{k+i-1}=t^k\sum_{j=0}^{n-k-1}{n-k-1\choose j}(-t)^{j}=t^k(1-t)^{n-k-1}
$$
hence
$$
S=n{n-1\choose k}\int_0^1t^k(1-t)^{n-k-1}\mathrm dt=n{n-1\choose k}\mathrm{B}(k+1,n-k)
$$
where the letter $\mathrm{B}$ refers to the beta numbers, defined as
$$
\mathrm{B}(i,j)=\frac{(i-1)!(j-1)!}{(i+j-1)!}
$$
All this yields
$$
S=n{n-1\choose k}\frac{k!(n-k-1)!}{n!}=1
$$
