Proving that $\sum_{j=0}^n(-1)^j\binom{n}{j} = \binom{n}{0} - \binom{n}{1} + ... \pm \binom{n}{n}=0$
Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$