An identity involving Stirling numbers of the second kind and binomial coefficients Need to prove:
$$\sum\limits_{k=0}^{n} \binom nk k^r x^k = \sum\limits_{j=0}^{r} \binom nj j! (1+x)^{n-j} x^j S(r,j)$$ 
where $S (n, k)$ denotes a Stirling number of second kind, the number of partitions of a set with $n$ elements into $k$ blocks.

This  sum can  also be  evaluated with  the technique  of annihilated
coefficient extractors (ACE).
Start with the generating function
$$F(w) = \sum_{r\ge 0} \frac{w^r}{r!} \sum_{j=0}^r {n\choose j} 
\times j! \times (1+x)^{n-j} \times x^j \times
{r\brace j}.$$
Recall the  bivariate generating function for the  Stirling numbers of
the second kind which is
$$G(z, u) = \exp(u(\exp(z)-1)).$$
Substitute this into the generating function to get
$$F(w) = \sum_{r\ge 0} \frac{w^r}{r!} \sum_{j=0}^r {n\choose j} 
\times j! \times (1+x)^{n-j} \times x^j \times
r! [z^r] \frac{(\exp(z)-1)^j}{j!}.$$
Switching summations we obtain
$$F(w) = \sum_{j\ge 0}  {n\choose j} 
\times (1+x)^{n-j} \times x^j \times
\sum_{r\ge j} w^r [z^r] (\exp(z)-1)^j.$$
The  inner  sum  contains  an  annihilated  coefficient  extractor  and
thus $F(w)$ simplifies to
$$F(w) = \sum_{j\ge 0}  {n\choose j} 
\times (1+x)^{n-j} \times x^j \times (\exp(w)-1)^j.$$
Apply the binomial theorem to get
$$F(w) = (x(\exp(w)-1) + 1 + x)^n = (1 + x\exp(w))^n.$$
Expand with the binomial theorem to obtain
$$F(w) = \sum_{k=0}^n {n\choose k} x^k \exp(kw).$$
Perform coefficient extraction to conclude:
$$r! [w^r] F(w) = r! [w^r] \sum_{k=0}^n {n\choose k} x^k \exp(kw)
= r! \times \sum_{k=0}^n {n\choose k} x^k \frac{k^r}{r!}
= \sum_{k=0}^n {n\choose k} x^k k^r.$$
The ACE technique is also used at this MSE link.
