Application to sums

Expansion. $(1-x)^n=\sum_k\binom{n}{k}(-x)^k$.

Derivative. $-n(1-x)^{n-1}=\sum_k\binom{n}{k}k(-x)^{k-1}(-1)$... at $x=1$, RHS $=\sum_k k\binom{n}{k}(-1)^k\cdot(-1)$? Let me just use the known formula.

Alternatively: $\sum_k\binom{n}{k}(-1)^k(n-k)=n\sum_k\binom{n}{k}(-1)^k-\sum_k k\binom{n}{k}(-1)^k=0-\sum_k k\binom{n}{k}(-1)^k$.

And $\sum_k k\binom{n}{k}(-1)^k=\frac{d}{dx}(1+x)^n\Big|_{x=-1}=n(1-1)^{n-1}=0$ for $n\ge2$.

Conclusion. For $n\ge2$, the sum is $0$. For $n=1$, the sum is $-1$.