TITLE: Proving $\sum_{i=1}^{n} {{n}\choose{i}} (-1)^i = -1$ by induction
QUESTION [5 upvotes]: I would like to prove this $$\sum_{i=1}^{n} {{n}\choose{i}} (-1)^i = -1$$ for all $n \in \mathbb{N} $. 

I started by replacing consecutively $n$ by $1, 2, 3$, one at a time, and verified that all of these indeed equal $-1$. However, I would like to prove it for all $i$, therefore I tried to do it by Mathematical Induction but failed somewhere. Below is my rationale.
$ n=1 \Rightarrow \sum_{i=1}^{n} {{n}\choose{i}} (-1)^i = \sum_{i=1}^{1} {{1}\choose{i}} (-1)^i = {{1}\choose{1}} (-1)^1  = 1 \times (-1) = -1$
$$\sum_{i=1}^{n+1} {{n+1}\choose{i}} (-1)^i \\= 
\sum_{i=1}^{n} {{n+1}\choose{i}} (-1)^i + \sum_{i=n+1}^{n+1} {{n+1}\choose{i}} (-1)^i \\
= \sum_{i=1}^{n} {{n+1}\choose{i}} (-1)^i + {{n+1}\choose{n+1}} (-1)^{n+1}\\
 = \sum_{i=1}^{n} \space [{{n}\choose{i-1}} + {{n}\choose{i}}] (-1)^i + {{n+1}\choose{n+1}} (-1)^{n+1} \\
= \sum_{i=1}^{n} \space {{n}\choose{i-1}} (-1)^i + \sum_{i=1}^{n}{{n}\choose{i}} (-1)^i + {{n+1}\choose{n+1}} (-1)^{n+1} = $$
Given the induction hypothesis, $\sum_{i=1}^{n} {{n}\choose{i}} (-1)^i = -1 $, hence we have
$ \sum_{i=1}^{n} \space {{n}\choose{i-1}} (-1)^i - 1 + (-1)^{n+1} = $
Now the way I see this is that $\sum_{i=1}^{n} \space {{n}\choose{i-1}} (-1)^i$ is similar to $\sum_{i=1}^{n}{{n}\choose{i}} (-1)^i $, but instead of having all the Pascal triangle terms except the term in ${n}\choose{0}$, we end up having all elements except the term in ${n}\choose{n}$. This is why I wrote the previous equation as equal to
$ \sum_{i=1}^{n} \space {{n}\choose{i}} (-1)^i + {{n}\choose{0}} (-1)^0 - {{n}\choose{n}} (-1)^n - 1 + (-1)^{n+1} = $
$ -1 + 1 - {{n}\choose{n}} (-1)^n - 1 + (-1)^{n+1} = - (-1)^n - 1 + (-1)^{n+1} = -1 + (-1)^{n+1} + (-1)^{n+1} = $
$ = -1 + 2 \times (-1)^{n+1} $
You see there is a problem in this resolution, because, depending on whether n is odd or even, the final result of this sum will differ. If $n$ is odd, $n+1$ is even, hence we have $-1 + 2 \times 1 = 1$. If $n$ is even, then $n+1$ is odd, hence we have $-1 + 2 \times (-1) = -3$ ! 
I did this resolution over and over, but I did not find the hole in my logic. Please help.
Thank you.

REPLY [2 votes]: \begin{align}
\sum_{i=1}^n\binom{n}i(-1)^i
&=\sum_{i=1}^n\left[\binom{n-1}{i}+\binom{n-1}{i-1}\right](-1)^i
\\&=\sum_{i=1}^n\left[\binom{n-1}{i}(-1)^i-\binom{n-1}{i-1}(-1)^{i-1}
\right]\end{align}
The last summation is telescoping. To see this, let $a_i=\binom{n-1}i(-1)^i$, and note the last sum is $\sum_{i=1}^n(a_i-a_{i-1}). $ The value of this sum is therefore $$a_n-a_0=\binom{n-1}n(-1)^n-\binom{n-1}0(-1)^0=0-1=-1.$$
More generally,
$$
\sum_{i=p}^q\binom{n}i(-1)^i=\binom{n-1}q(-1)^q-\binom{n-1}{p-1}(-1)^{p-1}.
$$