$$\sum\_{k=0}^n\cos(k\pi)\binom{n}{k}=0$$

I approached this problem having no idea that $\cos(k\pi)$ could be substituted so easily.
I tried first to expand the sum to $n$. So I wanted to ask if anyone knows how to expand this kind of sum? Would highly appreciate it. Is it like $1\times\cos(k\pi)$ and expand $(1+\cos(k\pi))^n$?