We have that $(x)_k - (x-1)_k = k (x-1)_{k-1}$. So applying the linear operator $f \mapsto xf(x) - xf(x-1)$, to the identity $$ \sum_{k=1}^{n} \genfrac\{\}{0pt}{}{n}{k}(x)_k = x^n $$ we get that $$\sum_{k = 1}^n \genfrac\{\}{0pt}{}{n}{k} k (x)_k = x^{n+1} - x(x-1)^n.$$
Edit: In retrospect, there is also a enumerative proof. The left hand side the set of functions from $[n] \to [x]$ together with a choice of point in the image. The right hand counts the set functions from $[n] \sqcup * \to [x]$ minus the set of functions such that the image of $*$ and $[n]$ are disjoint.
