Counting summation of binomial coefficients

This sum can be written as $$S=\sum_{k=0}^{n} (-1)^k (k+1){n \choose k}~~~~(1)$$The binomial theorem : $$\sum_{k=0}^{n} {n \choose k} (-x)^k=(1-x)^n~~~(2)$$ It gives $$\sum_{k=0}^{n}(-1)^k {n \choose k}=0 ~~~(3)$$ D (1) w.r. t $x$ and put $x=1$, we again get $$\sum_{k=0}^{n} (-1)^k k {n \choose k}=0~~~~(4)$$ From (3,4) it follows that $S=0$.