A sum involving Stirling numbers of the second kind For $p \in (0,1)$ and integers $n,k \geq 1$, what is known of the following sum?
$$
\sum_{j=1}^k j!S(k, j){n \choose j} p^j.
$$
Can we simplify it?
Edit: The question in this post: (A combinatorial sum and identity involving Stirling numbers of the second kind) gives a more general question and sum than given here, with the $p^j$ replaced by arbitrary $a(j)$. I thought the case $a(j) = p^j$ or $a(j) = x^j$ in general, is special enough to ask separately. I believe this particular case has connections with the moments of binomial random variables.
 A: Indeed, the sum is the $k$-th moment of the binomial random variable 
$$S_n = X_1 + \cdots + X_n, $$ 
where the $X_i$ are independent Bernoulli random variables, each with expected value $p$. 
We can compute this directly:
\begin{align}
\mathbb{E}[S_n^k] &= \mathbb{E}[\sum_{1 \leq m_1, \ldots, m_k \leq n} X_{m_1} \cdots X_{m_k}] = \sum_{1 \leq m_1, \ldots, m_k \leq n} \mathbb{E}[X_{m_1} \cdots X_{m_k}]\\
&= \sum_{j=1}^k\sum_{\substack{a_1, \ldots, a_j \geq 1\\ a_1 + \cdots + a_j = k}} {k \choose a_1, \ldots, a_j} \sum_{1 \leq m_1 < \cdots < m_j \leq n}\mathbb{E}[X_{m_1}^{a_1} \cdots X_{m_j}^{a_j}]\\
&= \sum_{j=1}^k\sum_{\substack{a_1, \ldots, a_j \geq 1\\ a_1 + \cdots + a_j = k}} {k \choose a_1, \ldots, a_j} \sum_{1 \leq m_1 < \cdots < m_j \leq n}\mathbb{E}[X_{m_1} \cdots X_{m_j}]\\
&= \sum_{j=1}^k j! S(k, j) \sum_{1 \leq m_1 < \cdots < m_j \leq n} \mathbb{E}[X_{m_1}]\cdots \mathbb{E}[X_{m_j}]\\
&= \sum_{j=1}^k j! S(k, j) \sum_{1 \leq m_1 < \cdots < m_j \leq n} p^j\\
&= \sum_{j=1}^k j! S(k, j) {n \choose j} p^j.
\end{align}
The $k$-th moment is the $k$-th derivative at $0$ of the moment generating function of $S_n$. Thus
$$
\sum_{j=1}^k j! S(k, j) {n \choose j} p^j = \left(\dfrac{d^k}{dt^k}(1-p + pe^t)^n\right)\mid_{t=0}.
$$
