Sum with Stirling numbers Show that for each $n>1$
$$\sum\limits_{k=1}^{n-1} \frac{(n-1)!}{(k-1)!}S(n,n-k) = (n-1)^n $$
where $S(n,m)$ is the Stirling number of the second kind. 
 A: Consider this:
$$ m^n = \sum_{j=1}^m  \binom{m}{j} \,   j! \, S(n,j)  $$ 
The left side is the number of ways of placing $n$ distinct balls in $m$ distinct cells. The right side is the same, each term corresponds to the number of configurations that have $j$ occupied (non-empty) cells. 
Then, replacing $k=m-j+1$:
$$ m^n =  \sum_{j=1}^m  \frac{m!}{(m-j)!} \, S(n,j) =  \sum_{k=1}^m  \frac{m!}{(k-1)!} \, S(n,m-k+1)$$
Pick the special case  $m=n-1$, and you get your identity.
A: Suppose we are trying to show that
$$\sum_{k=1}^{n-1} \frac{(n-1)!}{(k-1)!} {n\brace n-k}
= (n-1)^n.$$
Recall the species for set partitions which is
$$\mathfrak{P}(\mathcal{U} \mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1)).$$
This implies for the sum being evaluated that it is
$$\sum_{k=1}^{n-1} \frac{(n-1)!}{(k-1)!} 
\times n! [z^n] \frac{(\exp(z)-1)^{n-k}}{(n-k)!}.$$
This is
$$n! [z^n] \sum_{k=1}^{n-1} {n-1\choose k-1} (\exp(z)-1)^{n-k}
\\ = n! [z^n] (\exp(z)-1)^{n-1}
\sum_{k=1}^{n-1} {n-1\choose k-1} (\exp(z)-1)^{-(k-1)}
\\ =  n! [z^n] (\exp(z)-1)^{n-1} \times
\left(-(\exp(z)-1)^{-(n-1)}
+ \sum_{k=1}^n {n-1\choose k-1} (\exp(z)-1)^{-(k-1)} \right)
\\ =  n! [z^n] (\exp(z)-1)^{n-1} \times
\\ \left(-(\exp(z)-1)^{-(n-1)}
+ \left(1 + \frac{1}{\exp(z)-1}\right)^{n-1}\right)
\\ =  n! [z^n] (\exp(z)-1)^{n-1} \times
\\ \left(-(\exp(z)-1)^{-(n-1)}
+ \left(\frac{\exp(z)}{\exp(z)-1}\right)^{n-1}\right)
\\ = n! [z^n] (-1 + \exp(z(n-1)))
= (n-1)^n.$$
Addendum.
I just realized that starting from 
$$n! [z^n] \sum_{k=1}^{n-1} {n-1\choose k-1} (\exp(z)-1)^{n-k}$$
we can also continue with
$$n! [z^n] \sum_{k=1}^{n-1} {n-1\choose k-1} (\exp(z)-1)^{(n-1)-(k-1)}
\\ = n! [z^n]
\left(-1 + 
\sum_{k=1}^n {n-1\choose k-1} (\exp(z)-1)^{(n-1)-(k-1)}\right)
\\ = n! [z^n]
\left(-1 + (\exp(z)-1+1)^{n-1}\right)
\\= n! [z^n] \left(-1 + \exp(z(n-1))\right)
= (n-1)^n.$$
