Limit and convergence of $\frac{1}{n^{n-1}}\sum_{p=2}^{n-1} \left[ {n \choose p} (n-p)^{n-2} p (-1)^p \right]$ This is a part of larger question, in which I was asked to show that a certain ratio has a limit of $e^{-1}$.
After much of algebraic manipulation, I've found this ratio to be 
$$
\frac{1}{n^{n-1}}\sum_{p=2}^{n-1} \left[  {n \choose p} (n-p)^{n-2} p (-1)^p \right]
$$
and hopefully this is correct. Given that, this series must converge to $e^{-1}$ as $n\to\infty$, but I am stuck at showing this. 
So knowing what this series should converge to, I can rewrite the summand as
\begin{align*}
\frac{1}{n^{n-1}}\left[  {n \choose p} (n-p)^{n-2} p (-1)^p \right]
&=\frac{n}{n} \frac{n-1}{n} \cdots \frac{n-p+1}{n} \cdot \frac{1}{(p-1)!} \frac{(n-p)^{n-2}}{n^{n-p-1}} (-1)^p
\end{align*}
and a slight hope of improvement seems to emerge with factorial terms. But at this point I'm rather hopelessly lost, having no idea how to proceed. 
Obviously one approach would be bound the series by series that tend to $e^{-1}$, but I fail to spot any good candidate.
Any helps appreciated
 A: Assuming $n\geq 3$, we have:
$$\begin{eqnarray*}\sum_{p=2}^{n-1}\binom{n}{p}p(n-p)^{n-2}(-1)^p &=& \sum_{k=1}^{n-2}\binom{n}{k}(n-k)\,k^{n-2}(-1)^{n-k}\\&=&n\cdot\sum_{k=1}^{n-2}\binom{n-1}{k}k^{n-2}(-1)^{n-k}\tag{1}\end{eqnarray*} $$
but since:
$$ f(x)=\sum_{k=1}^{n-2}\binom{n-1}{k}e^{kx}(-1)^{n-k} = e^{(n-1)x}-(e^x-1)^{n-1}-(-1)^n \tag{2}$$
in order to compute $(1)$ we just need to compute $n\cdot\frac{d^{n-2}f}{dx^{n-2}}(0)$. 
Now look at the Wikipedia page about Stirling numbers of the second kind and notice that:
$$\lim_{n\to +\infty}\frac{n}{n^{n-1}}\cdot\left.\frac{d^{n-2}}{dx^{n-2}}(e^{(n-1)x})\right|_{x=0}=\lim_{n\to +\infty}\frac{(n-1)^{n-2}}{n^{n-2}}=\frac{1}{e}.\tag{3}$$
A: Suppose we seek to investigate
$$\frac{1}{n^{n-1}}
\sum_{p=2}^{n-1} {n\choose p} (-1)^p p (n-p)^{n-2}.$$
Taking $n\gt 2$ this becomes
$$\frac{1}{n^{n-1}}
\left(n\times (n-1)^{n-2}
+ \sum_{p=0}^{n} {n\choose p} (-1)^p p (n-p)^{n-2}\right).$$
The first term here is
$$\frac{(n-1)^{n-2}}{n^{n-2}}
= \left(1-\frac{1}{n}\right)^{n-2}
= \left(\frac{n}{n-1}\right)^2\left(1-\frac{1}{n}\right)^{n}.$$
This is $$\frac{1}{e}$$ in the limit by inspection.

The second term is the sum term which we now evaluate.
This is the sum:
$$\sum_{p=0}^{n} {n\choose p} (-1)^p p (n-p)^{n-2}.$$
Put $$(n-p)^{n-2}
= \frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp((n-p)z) \; dz.$$
This gives for the sum
$$\frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} 
\sum_{p=0}^{n} {n\choose p} (-1)^p p
\exp((n-p)z) \; dz
\\ = \frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp(nz)
\sum_{p=0}^{n} {n\choose p} (-1)^p p
\exp(-pz) \; dz.$$
We have
$$x((1+x)^n)' = x \sum_{q=1}^n {n\choose q} q x^{q-1}
= \sum_{q=0}^n {n\choose q} q x^{q}
= nx(1+x)^{n-1}.$$
Applying this to the sum
$$\frac{n \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp(nz)
(-\exp(-z)) (1-\exp(-z))^{n-1} \; dz
\\ = -\frac{n \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp((n-1)z)
(1-\exp(-z))^{n-1} \; dz
\\ = -\frac{n \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}}
(\exp(z)-1)^{n-1} \; dz
\\ = -\frac{n! \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}}
\frac{(\exp(z)-1)^{n-1}}{(n-1)!} \; dz
\\ = 
- n!\times (n-2)! \times {n-2\brace n-1} = 0.$$
The  sum term  evaluates to  zero leaving  only the  contribution from
$$\frac{1}{e}$$ which was to be shown.

Remark. If the Stirling number troubles anyone just observe that $\exp(z)-1$ starts at $z$ and therefore the power $z^{n-2}$ does not appear in $(\exp(z)-1)^{n-1}.$
