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.

We can use a generating-function approach.
For any $k$, define
$$
a_n=\sum_{i=1}^{n-k}(-1)^{i-1}\frac{(k+i-1)!}{k!\,(i-1)!}\frac{n!}{(k+i)!\,(n-k-i)!}\tag{1}
$$
Consider the sum
$$
\begin{align}
\sum_{n=k}^\infty a_nx^n
&=\sum_{n=k}^\infty\sum_{i=1}^{n-k}(-1)^{i-1}\frac{(k+i-1)!}{k!\,(i-1)!}\frac{n!}{(k+i)!\,(n-k-i)!}x^n\tag{2}\\
&=\sum_{i=1}^\infty\sum_{n=k+i}^\infty(-1)^{i-1}\binom{k+i-1}{k}\binom{n}{n-k-i}x^n\tag{3}\\
&=\sum_{i=1}^\infty\sum_{n=k+i}^\infty(-1)^{n-k-1}\binom{k+i-1}{k}\binom{-k-i-1}{n-k-i}x^n\tag{4}\\
&=\sum_{i=1}^\infty\sum_{n=0}^\infty(-1)^{n+i-1}\binom{k+i-1}{k}\binom{-k-i-1}{n}x^{n+k+i}\tag{5}\\
&=\sum_{i=1}^\infty(-1)^{i-1}\binom{k+i-1}{k}(1-x)^{-k-i-1}x^{k+i}\tag{6}\\
&=\sum_{i=0}^\infty(-1)^i\binom{k+i}{i}(1-x)^{-k-i-2}x^{k+i+1}\tag{7}\\
&=\frac{x^{k+1}}{(1-x)^{k+2}}\sum_{i=0}^\infty\binom{-k-1}{i}\left(\frac{x}{1-x}\right)^i\tag{8}\\
&=\frac{x^{k+1}}{(1-x)^{k+2}}(1-x)^{k+1}\tag{9}\\
&=\frac{x^{k+1}}{1-x}\tag{10}\\
&=\sum_{n=k+1}^\infty x^n\tag{11}
\end{align}
$$
Thus, for $n\gt k$, $a_n=1$.

Explanation
$(2)$  use $(1)$
$(3)$  change order of summation, rewrite binomials
$(4)$  rewrite negative binomial
$(5)$  substitute $n\mapsto n+k+i$
$(6)$  sum in $n$ using the binomial theorem
$(7)$  substitute $i\mapsto i+1$
$(8)$  collect terms, rewrite negative binomial
$(9)$  sum in $i$ using the binomial theorem
$(10)$ collect terms
$(11)$ rewrite function as a sum
