Suppose we seek to evaluate
$\sum_{k=0}^n (-1)^k {1+p+q\choose k} {p+n-k\choose n-k} {q+n-k\choose n-k}$ which is claimed to be ${p\choose n}{q\choose n}.$
Introduce ${p+n-k\choose n-k} = \frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{(1+z_1)^{p+n-k}}{z_1^{n-k+1}} \; dz_1$
and ${q+n-k\choose n-k} = \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{(1+z_2)^{q+n-k}}{z_2^{n-k+1}} \; dz_2.$
Observe that these integrals vanish when $k\gt n$ and we may extend $k$ to infinity.
We thus obtain for the sum $\frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{(1+z_1)^{p+n}}{z_1^{n+1}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{(1+z_2)^{q+n}}{z_2^{n+1}} \\ \times \sum_{k\ge 0} {1+p+q\choose k} (-1)^k \frac{z_1^k z_2^k}{(1+z_1)^k (1+z_2)^k} \; dz_2\; dz_1.$
This is $\frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{(1+z_1)^{p+n}}{z_1^{n+1}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{(1+z_2)^{q+n}}{z_2^{n+1}} \\ \times \left(1-\frac{z_1 z_2}{(1+z_1)(1+z_2)}\right)^{p+q+1} \; dz_2\; dz_1$
or $\frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{(1+z_1)^{n-q-1}}{z_1^{n+1}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{(1+z_2)^{n-p-1}}{z_2^{n+1}} (1+ z_1 + z_2)^{p+q+1} \; dz_2\; dz_1$
Supposing that $p\ge n$ and $q\ge n$ this may be re-written as $\frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{1}{z_2^{n+1} (1+z_2)^{p+1-n}} \\ \times (1+ z_1 + z_2)^{p+q+1} \; dz_2\; dz_1$
Put $z_2 = (1+z_1) z_3$ so that $dz_2 = (1+z_1) \; dz_3$ to get $\frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{1}{(1+z_1)^{n+1} z_3^{n+1} (1+(1+z_1)z_3)^{p+1-n}} \\ \times (1+ z_1)^{p+q+1} (1+z_3)^{p+q+1} \; (1+z_1) \; dz_3\; dz_1$
which is $\frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{(1+z_1)^p}{z_1^{n+1}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{1}{z_3^{n+1} (1+ z_3 + z_1 z_3)^{p+1-n}} \\ \times (1+z_3)^{p+q+1} \; dz_3\; dz_1 \\ = \frac{1}{2\pi i} \int_{|z_1|=\epsilon} \frac{(1+z_1)^p}{z_1^{n+1}} \frac{1}{2\pi i} \int_{|z_2|=\epsilon} \frac{(1+z_3)^{n+q}}{z_3^{n+1} (1 + z_1 z_3 /(1+z_3))^{p+1-n}} \; dz_3\; dz_1$
Extracting the residue for $z_1$ first we obtain $\sum_{k=0}^n {p\choose n-k} \frac{(1+z_3)^{n+q}}{z_3^{n+1}} {k+p-n\choose k} (-1)^k \frac{z_3^k}{(1+z_3)^k}.$
The residue for $z_3$ then yields $\sum_{k=0}^n (-1)^k {p\choose n-k} {k+p-n\choose k} {n-k+q\choose n-k}.$
The sum term here is $\frac{p!\times (p+k-n)!\times (q+n-k)!} {(n-k)! (p+k-n)! \times k! (p-n)! \times (n-k)! q!}$ which simplifies to $\frac{p!\times n! \times (q+n-k)!} {(n-k)! \times n!\times k! (p-n)! \times (n-k)! q!}$ which is ${n\choose k} {p\choose n}{q+n-k\choose q}$ so we have for the sum ${p\choose n} \sum_{k=0}^n {n\choose k} (-1)^k {q+n-k\choose q}.$
To evaluae the remaining sum we introduce ${q+n-k\choose q} = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{(1+v)^{q+n-k}}{v^{q+1}} \; dv$
getting for the sum ${p\choose n} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{(1+v)^{q+n}}{v^{q+1}} \sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+v)^k} \; dv \\ = {p\choose n} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{(1+v)^{q+n}}{v^{q+1}} \left(1-\frac{1}{1+v}\right)^n \; dv \\ = {p\choose n} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{(1+v)^{q}}{v^{q-n+1}} \; dv = {p\choose n} {q\choose q-n}$
which is ${p\choose n} {q\choose n}.$
This concludes the argument.