TITLE: Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$
QUESTION [13 upvotes]: Let $n$ be a positive integer. Prove that
  \begin{align}
\sum_{k=0}^n \left(-1\right)^k \binom{n}{k} = 0 .
\end{align}

I tried to solve it using induction, but that got me nowhere. I think the easiest way to prove it is to think of a finite set of $n$ elements, but I can't find the solution.

REPLY [2 votes]: Alternatively, prove a more general identity below: $$\sum_{r=0}^k\,(-1)^r\binom{n}{r}=(-1)^k\binom{n-1}{k}\,,$$
for all integers $n,k\geq 0$.
When $k=n$, we have 
$$\sum_{r=0}^{n}\,(-1)^r\binom{n}{r}=(-1)^n\binom{n-1}{n}=0\,.$$