We have that $(x)_k - (x-1)_k = k (x-1)_k$.   So applying the linear operator $f \mapsto xf(x) - xf(x-1)$, to the identity  $$ \sum_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(x)_k = x^n $$ we get that  $$\sum_{k = 1}^n \genfrac\{\}{0pt}{}{n}{k} k (x)_k = x^{n+1} - x(x-1)^n.$$