How can I prove that $\sum \limits_{i=0}^k {n \choose i}(-1)^i = {n-1 \choose k}(-1)^k$? I would like to prove this using a combinatorics argument. I have a feeling it involves the theorem $\sum \limits_{i=k}^n {i \choose k} = {n+1 \choose k+1}$. I'm not sure how to manipulate it to get the desired answer though. I also think I could explain it using Pascal's triangle, but I don't think that's a rigorous proof.
Thanks for any hints!

Hint: Use the fact that
$$ { n-1 \choose k} + {n-1 \choose k+1} = { n \choose k}.$$
