I want to show, using the Binomial Theorem, that
$\binom{n}{0} - \binom{n}{1} + \binom{n}{2} + ... + \left ( -1 \right )^{n}\binom{n}{n} = 0$
By using the symmetry of $\binom{n}{0} = \binom{n}{n}$ it is easy for all even $n$, but how does one show it when n is odd? The middle combination of the sequence must then cancel out all the others, right? Meaning
$-\binom{^{n}}{\frac{n+1}{2}}+2\sum_{k=0}^{\frac{n-1}{2}} \binom{n}{k}-1^{k}=0$
Surely there is a better way.