Let integers $n,k$ satisfy $0 \le k \le n$. We desire proof that $$ {n\choose k} = \sum {n\choose a}(-1)^a\;{-k\choose b}(-1)^b\;{-(n-k)\choose c}(-1)^c \tag{$*$}$$ where the (finite) sum is over all ordered triples $(a,b,c)$ of nonnegative integers satisfying $$ a\cdot n + b\cdot k + c\cdot(n-k) = k(n-k) \\ \text{(or equivalently}\quad \frac{a+b}{n-k} + \frac{a+c}{k} = 1 \quad). $$ Of course the binomial coefficient for a binomial to a negative power is $$ {-k\choose b} = \frac{(-k)(-k-1)(-k-2)\cdots(-k-b+1)}{b!} . $$

Now two of the terms in ($*$) are $(a,b,c)=(0,n-k,0)$ and $(a,b,c)=(0,0,k)$. Those two terms do, indeed, add to ${n\choose k}$. But then the problem becomes: show that the rest of the terms sum to zero.

**Note**

This arose from my attempt to solve
integral of a "sin-omial" coefficients=binomial .
It is the calculation of the residue at $w=0$ of rational function
$$
\frac{(w^n-w^{-n})^n}{(w^k-w^{-k})^k(w^{n-k}-w^{-(n-k)})^{n-k}w}
$$
obtained by changing variables in the integral of that problem.