Let $n, k \geq 1$. Let $a(j),\, 1\leq j \leq k$, be a sequence of real numbers. Consider the sum $$ \sum_{j=1}^k j! S(k, j) {n \choose j} a(j), $$ where $S(k,j)$ are Stirling numbers of the second kind. Is there some kind of "inversion" identity that simplifies the sum? What is the sum in the simple case that $a(j) = 1$ for each $j$? Thanks.
Update: In the case when the $a(j) = 1$ the sum is well-known. It equals $n^k$. See Marc van Leeuwen's answer in abiessu's question, and Brian M. Scott's answer below. Any words on $a(j)$ arbitrary?
The expression has a simple combinatorial interpretation: for a fixed $j$ the expression $j!S(k,j)\binom{n}{j}$ is the number of functions from $[k]$ to $[n]$ whose range has cardinality $j$, so the sum is just the number of functions from $[k]$ to $[n]$, which is $n^k$.