Alternate proof of $\sum_{j=0}^{m}\sum_{k=0}^{n}\left(-1\right)^{j+k}\binom{j+k}{k}\binom{m}{j}\binom{n}{k}=\delta_{mn}$

Question

A recent question used this identity. I'd like to know how to derive it other than by a proof I'll give here. It uses$$\begin{align}\left(-1\right)^{j}\binom{m}{j} &=\left[x^{j}\right]\left(1-x\right)^{m},\\ \left(-1\right)^{k}\binom{j+k}{k}&=\left[y^{k}\right]\left(1-y\right)^{j+k},\\ \binom{n}{k}&=\left[z^{k}\right]\left(1+z\right)^{n}.\end{align}$$(I suspect these factors can be taken as the entries of three matrices in a different proof strategy, but for now I'll discuss what happens when we rewrite them as power series coefficients.) Since $\sum_{l}v^{l}\left[u^{l}\right]f\left(u\right)=f\left(v\right)$ for polynomial $f$,$$\begin{align}\delta_{mn}&=\left[x^{m}\right]x^{n} \\&=\sum_{k}\left[x^{m}\right]\left(x-1\right)^{k}\binom{n}{k} \\&=\sum_{k}\left(-1\right)^{k-m}\binom{k}{k-m}\binom{n}{k} \\&=\sum_{k}\left[y^{k-m}\right]\left(1-y\right)^{k}\left[z^{k}\right]\left(1+z\right)^{n} \\&=\sum_{k}\left[y^{k}\right]\left(1-y\right)^{k}\left[z^{k}\right]\left(1+z\right)^{n}y^{m} \\&=\sum_{k}\left[y^{k}\right]\left(1-y\right)^{k}\left[z^{k}\right]\left(1+z\right)^{n}\sum_{j}\left(1-y\right)^{j}\left[x^{j}\right]\left(1-x\right)^{m} \\&=\sum_{jk}\left[x^{j}\right]\left(1-x\right)^{m}\left[y^{k}\right]\left(1-y\right)^{j+k}\left[z^{k}\right]\left(1+z\right)^{n} \\&=\sum_{j=0}^{m}\sum_{k=0}^{n}\left(-1\right)^{j+k}\binom{j+k}{k}\binom{m}{j}\binom{n}{k}.\end{align}$$

Answer 1

From the integral representation of the binomial coefficient $$\dbinom{n}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{k+1}}dz$$ we have

\begin{align} \sum_{j=0}^{m}\sum_{k=0}^{n}\left(-1\right)^{j+k}\dbinom{j+k}{k}\dbinom{m}{j}\dbinom{n}{k} & =\frac{1}{2\pi i}\oint_{\left|z\right|=1}\sum_{j=0}^{m}\sum_{k=0}^{n}\left(-1\right)^{j+k}\dbinom{m}{j}\dbinom{n}{k}\frac{\left(1+z\right)^{k+j}dz}{z^{k+1}} \\ = & \frac{\left(-1\right)^{m+n}}{2\pi i}\oint_{\left|z\right|=1}\frac{z^{m}}{z^{n+1}}dz=\delta_{mn} \end{align} since, for $h\in\mathbb{Z}$, we have $$\oint_{\left|z\right|=1}z^{h}dz=i\int_{0}^{2\pi}e^{i\left(h+1\right)\theta}d\theta=\begin{cases} 2\pi i, & h=-1\\ 0, & \mathrm{otherwise.} \end{cases}$$