Sum of $k {n \choose k}$ is $n2^{n-1}$
Just change the index $s=k-1$
$$\sum_{k=1}^{n}k {n \choose k} = n\sum_{k=1}^{n} {n-1\choose k-1}= n\sum_{s=0}^{n-1} {n-1\choose s} =n2^{n-1}$$
One possible proof:
Because $$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k,$$ then by taking derivative, $$n(1+x)^{n-1} = \sum_{k=1}^n k{n \choose k} x^{k-1}. $$ Let $x=1$, we obtain that $$n2^{n-1} = \sum_{k=1}^n k {n \choose k}.$$