How to prove that $\sum_{j=0}^{k}{(-1)^j\binom{n}{j}}=(-1)^k\binom{n-1}{k}$?

Question

How to prove that $\sum_{j=0}^{k}{(-1)^j\binom{n}{j}}=(-1)^k\binom{n-1}{k}$?

LHS$=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}...(-1)^k\binom{n}{k}$

$=\frac12\left(\binom{n}{0}-\binom{n}{1}+\binom{n}{2}...(-1)^k\binom{n}{k}+(-1)^{k}\binom{n}{n-k}...+ \binom{n}{n-2}-\binom{n}{n-1}+\binom{n}{n}\right)$

Is this right? How to proceed next?