While trying to prove the power series expansion of $(1-z)^{-m}$ “by hand”, I'm stumped trying to prove the following identity
$$\sum_{k=0}^m (-1)^k {{(m+n-k-1)!} \over {k!\, (n-k)! \,(m-k)!}}=0,$$ where all the variables are non-negative integers and $n\ge m$.
The expression above is almost a multinomial coefficient. But I cannot make headway based on that knowledge.
I would appreciate any hints or references to literature on how to proceed to prove this.
This is not a full answer to your question, so forgive me if it is off-topic and does not help.
I computed $$S_p=\sum_{k=0}^p (-1)^k {{(m+n-k-1)!} \over {k!\, (n-k)! \,(m-k)!}}$$ and obtained after simplifications $$S_p=-\frac{(-1)^p (p+1) (-m-n+p+1) (m+n-p-2)!}{m n (p+1)! (m-p-1)! (n-p-1)!}$$ which can rewrite $$S_p=\frac{(-1)^p \Gamma (m+n-p)}{m n \Gamma (p+1) \Gamma (m-p) \Gamma (n-p)}$$ and which effectively cancels if $p=m$ because of the $\Gamma(m-p)$ term.
I hope and wish this gives you some ideas for a full proof.