Binomial Alternating Sum starting at index 1

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 $(2)$.

More generally $\displaystyle\;\sum_{i=0}^n a_i = a_0 + \sum_{\color{red}{i=1}}^n a_i\,$.
In this case $\displaystyle\;a_i = (-1)^i \binom{n}{i}\,$, therefore $\,a_0=1\,$, then $\displaystyle\;\sum_{i=0}^n (-1)^i \binom{n}{i} = 1 + \sum_{\color{red}{i=1}}^n (-1)^i \binom{n}{i}\,$.
It follows by the binomial theorem that $\displaystyle\;\sum_{\color{red}{i=1}}^n (-1)^i \binom{n}{i}=-1 + \sum_{i=0}^n (-1)^i \binom{n}{i} = -1 + (1-1)^n \,$.
• Nice answer, and thanks for making it clear where you got the $(1-1)^n$ term! – Linus Rastegar Sep 1 '17 at 8:24
The first it's Its $$(1-1)^n=0$$ The second it's $$(1-1)^n-1=-1$$
$$(1-1)^n=\binom{n}{0}1^n(-1)^0+\binom{n}{1}1^{n-1}(-1)^{1}++\binom{n}{2}1^{n-2}(-1)^{2}+...++\binom{n}{n}1^{n-n}(-1)^{n}=0^n=0\\\to \\\sum_{i=0}^n (-1)^i \binom{n}{i} = 0.$$now expand sigma for $i=0$ and you will have $$\sum_{i=0}^n (-1)^i \binom{n}{i} =\binom{n}{0}1^n(-1)^0+\sum_{i=1}^n (-1)^i \binom{n}{i} = 0\\1+\sum_{i=1}^n (-1)^i \binom{n}{i} = 0 \to \\\sum_{i=1}^n (-1)^i \binom{n}{i} =0-1$$