I recently came across the following sum on Wikipedia:
$$\sum\_{i=0}^n (-1)^i \binom{n}{i} = 0. \qquad (1)$$
I was wondering what proof exists for $$\sum\_{i=1}^n (-1)^i \binom{n}{i} = -1, \quad (2)$$ 
without assuming $(1)$. So far, I've split up the cases $n$ even/$n$ odd and have been able to prove that, when $n$ is odd, all terms except the last (which simplifies to $-1$) cancel. With $n$ even the cancellations are not as easy or obvious, though I've checked for a few even values of $n$ that $(2)$ still turns out to be $-1$. I'd appreciate a proof that shows that verifies $(2)$.