Is $\sum_{k=0}^{r}\binom{r}{k}(r-k-1)^{r-k}(k-1)^{k-1}+(r-2)^r=0$ right? How to prove $$\sum_{k=0}^{r}\binom{r}{k}(r-k-1)^{r-k}(k-1)^{k-1}+(r-2)^r=0$$
I met this function when I tried to give another proof of the known lower bound of Tur\'an functions of complete hypergraphs ( Based on a same construction, instead of using shifting method, I tried to count edges directly )
Here is the question:
Define $a_0=-1$ and $a_1=1$. For all $r\geqslant2$, $a_0,a_1,\cdots,a_r$ satisfy
$$\sum_{k=0}^{r}\binom{r}{k}(r-k-1)^{r-k}a_k+(r-2)^r=0.$$
Prove that $a_k=(k-1)^{k-1}$ for all $k\geqslant2$.
I've tried generating function (exponential form) and Stirling's inversion formula, but they seemed to be wrong directions. And I've verified it for some numbers and they all turned to be right. 
 A: Suppose we seek to evaluate
$$Q_r = \sum_{k=0}^r {r\choose k}
(r-k-1)^{r-k} (k-1)^{k-1}.$$
Concerning the exponential generating function for this quantity
$$Q(z) = \sum_{r\ge 0} Q_r \frac{z^r}{r!}$$
we observe that when  we multiply two exponential generating functions
of the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the  product of  the two generating  functions is  the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

Therefore what  we have  here is a  convolution of the  two generating
functions
$$A(z) = \sum_{r\ge 0} (r-1)^r \frac{z^r}{r!}
\quad\text{and}\quad
B(z) = \sum_{r\ge 0} (r-1)^{r-1} \frac{z^r}{r!}.$$

The species of labelled trees has the specification
$$\mathcal{T} =
\mathcal{Z} \times \mathfrak{P}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$

Now $A(z)$ counts endofunctions with no fixed points
which gives the species
$$\mathfrak{P}(\mathfrak{C}_{\ge 2}(\mathcal{T}))$$
which produces the generating function
$$A(z) = \exp\left(-T(z) + \log\frac{1}{1-T(z)}\right)
= \exp(-T(z)) \frac{1}{1-T(z)}
\\ = \frac{z}{T(z)} \frac{1}{1-T(z)}.$$
Observe that the generating function of unmodified endofunctions
is $$E(z) = \sum_{r\ge 0} r^r \frac{z^r}{r!}
= \exp\left(\log\frac{1}{1-T(z)}\right)
= \frac{1}{1-T(z)}.$$
We need to integrate this to obtain $B(z),$ getting
$$\int \frac{1}{1-T(z)} dz$$
Put $T(z) = w$ to  get $z=w\exp(-w)$ and $dz = (\exp(-w)-w\exp(-w)) \;
dw$ to obtain
$$\int \frac{1}{1-w} (1-w) \exp(-w) \; dw
\\ = - \exp(-w)
= - \frac{1}{w} w \exp(-w)
= - \frac{z}{T(z)}.$$
We get for the value at zero by L'Hopital
$$- \frac{1}{T'(0)} = -1$$
which means that we have the right constant.

Closing in we obtain
$$Q_r
= -\frac{r!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}}
\frac{z}{T(z)} \frac{z}{T(z)} \frac{1}{1-T(z)} \; dz.$$
Using the same substitution as before this becomes
$$-\frac{r!}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(r+1))}{w^{r+1}}
\exp(-2w) \frac{1}{1-w}
(1-w) \exp(-w) \; dw
\\ = -\frac{r!}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(r-2))}{w^{r+1}}
\; dw
= -(r-2)^r$$
as claimed.

Additional material on endofunctions and the labeled tree function may
be found at this
MSE link
and this
MSE link II.
A: One may recall Abel's identity:

$$\sum_{k=0}^n \binom{n}{k} (s-k)^{n-k} (\nu+k)^{k-1} = \frac{(\nu+s)^n}\nu, \qquad \nu\neq0.\tag1$$

A short WZ-style proof of $(1)$ may be found here or here (Shalosh B. Ekhad and John E. Majewicz).
Then just apply $(1)$ with $s=r-1$, $\nu=-1$ to get your initial identity.
