I am looking for closed forms, or at least a good approximation for


$$f(n) = \sum\_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(n)\_kk$$


I know that


$$\sum\_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(n)\_k = n^n$$


I have the intuition that $f(n)$ is bounded above by $n^{n+1}$ and approaches $n^{n+1}$ for large $n$ but I am not entirely sure and don't know how to form a proof (or anti-proof).


Sorry if this question is too basic for math overflow, I wasn't sure if it belonged here or elsewhere.



Sum of the Stirling numbers of the second kind multiplied by $k$ and falling factorials
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 counts the set of functions from $[n] \to [x]$ together with a choice of point in the image. The right hand counts the set offunctions from $[n] \sqcup \* \to [x]$ minus the set of functions such that the image of $\*$ and $[n]$ are disjoint.


