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.
 A: I claim first that
$$\binom{n}kk^r=\sum_{i=\max\{0,k-r\}}^k\binom{n}{i,k-i,n-k}{r\brace{k-i}}(k-i)!\;.\tag{1}$$
The lefthand side of $(1)$ clearly counts the ways to choose $K\subseteq[n]$ such that $|K|=k$ and then choose a function from $[r]$ to $K$. The $i$ term on the righthand side of $(1)$ counts the ways to choose a $k$-element subset $K$ of $[n]$, choose an $i$-element subset $I$ of $K$, and then choose a function from $[r]$ onto $K\setminus I$. Clearly this is possible if and only if $\max\{0,k-r\}\le i\le k$, so the two sides of $(1)$ count the same thing.
Now just rearrange the righthand side of the desired identity, expanding $(1+x)^{n-j}$ by the binomial theorem, making a change of index ($k=i+j$), and reversing the order of summation:
$$\begin{align*}
\sum_{j=0}^r\binom{n}j{r\brace j}j!(1+x)^{n-j}x^j&=\sum_{j=0}^r\binom{n}j{r\brace j}j!x^j\sum_{i=0}^{n-j}\binom{n-j}ix^i\\\\
&=\sum_{i=0}^n\sum_{j=0}^{\min\{r,n-i\}}\binom{n}j\binom{n-j}i{r\brace j}j!x^{i+j}\\\\
&=\sum_{i=0}^n\sum_{k=i}^{\min\{i+r,n\}}\binom{n}{k-i}\binom{n-k+i}i{r\brace{k-i}}(k-i)!x^k\\\\
&=\sum_{k=0}^n\sum_{i=\max\{0,k-r\}}^k\binom{n}{i,k-i,n-k}{r\brace{k-i}}(k-i)!x^k\;.
\end{align*}$$
Finally, apply $(1)$.
A: 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.
A: If you denote the operator of differentiating and multiplying by $x$ as $D_{x}$
Then we have that
$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{(k)}(x) x^{k}$$
where $s(n,k)$ is the stirling number of the second kind and $f^{(k)}(x)$ is the $k^{th}$ derivative of $f(x)$. 
This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$
Your identity is just $D_x$ applied to $$\frac{(1+x)^n}{n!}$$ $r$ times. Your $S(r,j)$ is same as $s(r,j)$ above.
A: Here is a combinatorial argument. Let us use the same notation for a natural number $x$ and for a set with $x$ elements.
On the left, you have cardinality of the set of data $(k\subseteq n, r\to k\to x)$ (meaning a $k$-element subset of $n$, a map from $r$ to this subset, and a map from this subset to $x$).
On the right, you have cardinality of the set of data $(j\subseteq n, r\twoheadrightarrow j, j\to x,(n-j)\to(1+x))$, (meaning a $j$-element subset of $n$, a map from $r$ onto this subset, a map from this subset to $x$, and another map from the complement of this subset to $1+x$ (this is clear - maybe just additionally noting that the number of onto maps from $r$ to $j$ equals $j!S(r,j)$)).
Now it is almost obvious how to establish a one-to-one correspondence between these kinds of data. Given $(k\subseteq n, r\to k\to x)$, let $j$ be the image of $r\to k$, let $j\to x$ be the restriction of $k\to x$ to $j\subseteq k$, and let $(n-j)\to(1+x)$ be given by sending $k-j$ to $x$ according to the given $k\to x$, and the remaining $n-k$ to $1\in(1+x)$. Conversely given data of the second kind, let $k$ be the (disjoint) union of $j$ and the inverse image of $x$ under $(n-j)\to(1+x)$, let $r\to k$ be the composite $r\twoheadrightarrow j\hookrightarrow k$, and let $k\to x$ be as $j\to x$ on $j$ and as $(n-j)\to(1+x)$ on the rest of $k$.
