Why does $\sum_{k=0}^{n} (-1)^k {n \choose k} = 0$ if $n\neq 0$ and 1 otherwise?

Question

Basically, I've computed a few values by hand, and those followed the pattern in the question. Wolframalpha also claims that this function equals the Kronecker delta function which follows $$\delta_{ij} = \begin{cases}0 &\text{if } i \neq j \\1 &\text{if }x=1\end{cases}$$ Why is this the case?

Answer 1

This counts the number of even subsets of a set of size $n$ minus the number of odd subsets. If $n=0$ there is only one subset, which is even ($\varnothing$); otherwise there are the same number of each (swapping whether or not $1$ is included gives a bijection from even to odd).

Answer 2

Answering my own question:

This is simply a special case of the equality

$$(a+b)^n = \sum_{k=0}^n {n \choose k}a^{n-k}b^k$$

with $a=1$ and $b=-1$.

Answer 3

Recall the binomial formula: $$(x+y)^n=\sum_{k=0}^n \binom nk x^k y^{n-k}$$

Then: $$0=0^n=((-1)+1)^n=\sum_{k=0}^n\binom nk (-1)^k (1)^k=\sum_{k=0}^n\binom nk (-1)^k$$