By the conclusion of # 3858, $$k\binom{n}{k}=n\binom{n-1}{k-1}$$ for $k\ge 1$. Therefore, $$\sum_{k=0}^{n}(-1)^{k}k\binom{n}{k} = 0 + \sum_{k=1}^{n}(-1)^{k}n\binom{n-1}{k-1}= n\sum_{k=0}^{n-1}(-1)^{k-1}\binom{n-1}{k}=\boxed{0}$$

The last step utilizes the conclusion of # 3158 by replacing $n$ with $(n-1)$.

An alternative solution is to take derivative of the identity $$(1-x)^n=\sum_{k=0}^n(-1)^k\binom{n}{k}x^k \implies n(1-x)^{n-1}=\sum_{k=0}^n(-1)^kk\binom{n}{k}x^{k-1}$$

Afterwards, letting $x=1$ gives the same result.