Binomial sum formula for $(n+1)^{n-1}$ Has anybody seen a proof for
$$
(n+1)^{n-1}=\frac{1}{2^n}\sum_{k=0}^n
C_n^k(2k+1)^{k-1}(2(n-k)+1)^{n-k-1} ?
$$
There are lots of reasons to think that this is true. In particular the formula holds for $n=0,1,2,3,4,5$.
 A: A (not straight) proof. 
Using Bürmann-Lagrange formula we expand the regular at $z=0$ solution $w(z)$ of the equation $w=ze^{aw^2}$ as the following series
$$
w(z)=\sum_{n=0}^{\infty}\frac{a^n(2n+1)^{n-1}} {n!}z^{2n+1}
$$
(see problem 26.07 in СБОРНИК ЗАДАЧ ПО ТЕОРИИ АНАЛИТИЧЕСКИХ ФУНКЦИЙ, Под редакцией М. А ЕВГРАФОВА (Russian), -- any other reference - for this particular expansion, not for  Bürmann-Lagrange formula - will be greatly appreciated). Using Cauchy formulas for the coefficients of the series products, we get from this expansion 
$$
w^2=\sum_{n=0}^{\infty}\frac{a^{n}} {n!}z^{2n+2} \sum_{k=0}^{n}C_n^k(2k+1)^{k-1}(2(n-k)+1)^{n-k-1}.
$$
On the other hand, since $w^2=z^2e^{2aw^2}$, applying the expansion in the problem 26.06 (from the same problem-book), we get that 
$$
w^2=\sum_{n=0}^{\infty}\frac{a^{n}2^n(n+1)^{n-1}} {n!}z^{2n+2}.
$$ 
Equating to each other the coefficients for $z^{2n+2}$ in both series we obtain the desired equality.
The first mentioned problem 26.07 is for any $m\geq1$ and states that the regular at $z=0$ solution $w(z)$ of the equation $w=ze^{aw^m}$ has the expansion
$$
w(z)=\sum_{n=0}^{\infty}\frac{a^n(mn+1)^{n-1}} {n!}z^{mn+1}
$$ 
(above we have used it only for the value $m=2$).
Based on the same two mentioned problems, the formula can be generalized for any $m\in\Bbb{N}$ as follows. 
Let $\alpha_k^{n}\in\Bbb{N}\cup\{0\};  \;\;    0\leq\alpha_k^{n}\leq n;   \;\; 0\leq k\leq m.$ Denote by $A_{m}^{n}:   
    =\{\alpha_1^{n},\alpha_2^{n},\ldots,\alpha_m^{n}\}$ an $n$-tuple of such numbers such that $\sum_{k=0}^m\alpha_k^{n}=n$. Further denote by $ (A_{m}^{n})!:=\alpha_1^{n}!\alpha_2^{n}!\cdots     \alpha_m^{n}!$, and by $ C_n^{A_{m}^{n}}:=\binom{{n}}{A_m^n}    := \frac{n!}{(A_{m}^{n})!}.$ Then,
\begin{gather*} 
\boxed{\quad(n+1)^{n-1}=\frac{1}{m^n}  \sum_{A_{m}^{n}}
 C_n^{A_{m}^{n}}   (m\alpha_1^{n}+1)^{\alpha_1^{n}-1}
 \cdots (m\alpha_m^{n}+1)^{\alpha_m^{n}-1}\quad}.
\end{gather*}
A: This is a particular case (for $x = 1$, $y = 1$ and $z = -2$) of the following result:

Let $n$ be a nonnegative integer. In the polynomial ring
  $\mathbb{Z}\left[  x,y,z\right]  $, we have
  $$
\sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(
n-k\right)  z\right)  ^{n-k-1}=\left(  x+y\right)  \left(  x+y-nz\right)
^{n-1}.
$$

This result is Theorem 2 in MathOverflow post #273459 (where I derive it from the famous Abel identity).
