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 $n=0,1,2,3,4,5$.

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).
