Sums of binomial coefficients Does anyone know something about the following sums?
$$
S_m(n)=\sum\limits_{k=o}^n(-1)^k{mn\choose mk}
$$
Notice that $S_m(n)=0$ for odd $n$, so we only consider $S_m(2n)$. It holds that $S_0(2n)=1$, $S_1(2n)=0$, $S_2(2n)=(-4)^n$, $S_3(2n)=\frac{2}{3}(-27)^n$, but $S_4(2n)$ is no longer a geometric progression.

Start by restating the problem: we seek to evaluate
$$S_m(n) = \sum_{k=0}^n (-1)^k {nm\choose km}
= \sum_{k=0}^n (-1)^k {nm\choose nm-km}.$$
Introduce the integral representation
$${nm\choose nm-km}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm-km+1}} \; dz.$$
This gives the following for the sum:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\sum_{k=0}^n \frac{(-1)^k}{z^{-km}} \; dz.$$
We  may extend  the sum  to  infinity because  the integral  correctly
represents the binomial coefficient being zero for $k>n$ to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\sum_{k\ge 0} \frac{(-1)^k}{z^{-km}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\sum_{k\ge 0} (-1)^k z^{km} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\frac{1}{1+z^m}\; dz.$$
Extracting coefficients we obtain
$$[z^{nm}]  (1+z)^{nm} \frac{1}{1+z^m}
= \sum_{q=0}^{nm} {nm\choose nm-q} [z^q] \frac{1}{1+z^m}
= \sum_{q=0}^{nm} {nm\choose q} [z^q] \frac{1}{1+z^m}.$$
Introduce  $\rho_k =  e^{i\pi/m +  2\pi ik/m}$  so that  using partial
fractions by residues on simple poles we have
$$\frac{1}{1+z^m}
= \sum_{k=0}^{m-1} 
\frac{\mathrm{Res}\left(\frac{1}{1+z^m}; z=\rho_k\right)}{z-\rho_k}.$$
This is
$$\sum_{k=0}^{m-1} 
\frac{\frac{1}{m} \rho_k^{-(m-1)}}{z-\rho_k}
= \sum_{k=0}^{m-1} 
\frac{\frac{1}{m} \rho_k^{-m}}{z/\rho_k-1}.$$
Substituting this into the sum formula yields
$$\sum_{q=0}^{nm} {nm\choose q} [z^q]
\sum_{k=0}^{m-1} 
\frac{\frac{1}{m} \rho_k^{-m}}{z/\rho_k-1}
= - \sum_{q=0}^{nm} {nm\choose q} 
\sum_{k=0}^{m-1} \frac{1}{m} \rho_k^{-m} \rho_k^{-q}$$
Now $\rho_k^{-m} = -1$ so this yields
$$\sum_{q=0}^{nm} {nm\choose q} 
\sum_{k=0}^{m-1} \frac{1}{m} \rho_k^{-q}
= \frac{1}{m} \sum_{k=0}^{m-1}  
\sum_{q=0}^{nm} {nm\choose q} \rho_k^{-q}
\\ = \frac{1}{m} \sum_{k=0}^{m-1}  
\left(1 + \frac{1}{\rho_k}\right)^{nm}
= \frac{1}{m} \sum_{k=0}^{m-1}  
\left(1 + \rho_k\right)^{nm}.$$

A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
