Sum involving Stirling numbers of the second kind Is it possible to simplify this sum:
$\sum\limits_{k=1}^{n} {n\brace k} (x)_k k =?$
Where, $ {n\brace k}$ are Stirling numbers of the second kind and $(x)_k$ if a falling factorial.
Note: it is well known that:
$\sum\limits_{k=1}^{n} {n\brace k} (x)_k  = x^n$
 A: By way of enrichment here is a proof using generating functions.
Suppose we seek to evaluate
$$\sum_{k=0}^n {n\brace k} k q^{\underline{k}}
= \sum_{k=0}^n {n\brace k} k {q\choose k} k!$$
with $q$ a positive integer.

The combinatorial class of set partitions marked by the number of parts is $$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}(\mathcal{U}\times \textsc{SET}_{\ge 1}(\mathcal{Z})).$$
This gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1))$$
which immediately yields
$${n\brace k} =
n! [z^n] \frac{1}{k!} \left(\exp(z)-1\right)^k.$$
This gives for the sum 
$$n! [z^n] \sum_{k=0}^n k {q\choose k} k!
\frac{1}{k!} \left(\exp(z)-1\right)^k
\\ = n! [z^n] \sum_{k=0}^n k {q\choose k}
\left(\exp(z)-1\right)^k.$$
Observe that
$$((1+x)^n)' = n(1+x)^{n-1} = 
\left(\sum_{p=0}^n {n\choose p} x^p\right)'
= \sum_{p=1}^n p {n\choose p} x^{p-1}$$
so that
$$nx (1+x)^{n-1} = \sum_{p=0}^n p {n\choose p} x^{p}.$$
There are two cases, first when $q\le n$ which gives
$$n! [z^n] \sum_{k=0}^q k {q\choose k}
\left(\exp(z)-1\right)^k$$
or $$n! [z^n] q (\exp(z)-1) \exp(z(q-1))$$
The second case is when $q\gt  n.$ But $\exp(z)-1$ starts at $z$ so we
can extend the summation from  $n$ to $q$ because the additional terms
do not contribute, again getting
$$n! [z^n] q (\exp(z)-1) \exp(z(q-1))$$
This is
$$n! [z^n] q (\exp(zq)-\exp(z(q-1)))
= q \times q^n - q \times (q-1)^n
= q^{n+1} - q (q-1)^n.$$
A: We can take advantage of the recurrence
$${{n+1}\brace k}=k{n\brace k}+{n\brace{k-1}}$$
to get
$$\begin{align*}
\sum_{k=1}^nk{n\brace k}x^{\underline k}&=\sum_{k=1}^n\left({{n+1}\brace k}-{n\brace{k-1}}\right)x^{\underline k}\\\\
&=\sum_{k=1}^n{{n+1}\brace k}x^{\underline k}-\sum_{k=1}^n{n\brace{k-1}}x^{\underline k}\\\\
&=x^{n+1}-x^{\underline{n+1}}-x\sum_{k=0}^{n-1}{n\brace k}(x-1)^{\underline k}\\\\
&=x^{n+1}-x^{\underline{n+1}}-x\big((x-1)^n-(x-1)^{\underline n}\big)\\\\
&=x^{n+1}-x(x-1)^n\;.
\end{align*}$$
