Product identity for $n^n$ I came across the rather nice identity
\begin{align}
&&\frac{(-n)^{n-1} \Gamma (n+1)}{(1-n)_{n-1}}&&\tag{1}&\\
\\
&=&\prod _{k=1}^{n-1} \frac{(k+1) n^2}{n^2-k n}&&\tag{2}&\\
\\
&=&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n}
\cdots\frac{n^3- 2n^2}{3 n}\cdot\frac{n^3- n^2}{2 n}\cdot n^2&&\tag{3}&\\
\\
&=&\underbrace{n\cdot n\cdot n\cdots n\cdot n\cdot n}_{n\text{ times}}&&\tag{4}&\\
\\
&=&n^{n}&&\tag{5}&\\
\end{align}
where ${(1-n)_{n-1}}$ in $(1) $ is the Pochhammer symbol, where $(x)_{n}\equiv\dfrac{\Gamma(x+n)}{\Gamma(x)}$. 
Interestingly, Mathematica doesn't simplify the expression to steps $(4)$ & $(5).$ How is this identity proven, and for what $n$ is this not the case?

Steps (1) to (2)
\begin{align}
f(n) 
&= \frac{(-n)^{n-1} \Gamma(n+1)}{(1-n)_{n-1}} \\
&= \frac{(-n)^{n-1} n!}
{\underbrace{(1-n)(1-n+1)(1-n+2)\cdots -2 \cdot -1}_{n-1 \tiny \mbox{ factors}}} \\
&= \frac{(-n)^{n-1} n!}
{(1-n)(2-n)(3-n)\cdots -2 \cdot -1} \\
&= \frac{n^{n-1} n!}
{(n-1)(n-2)(n-3)\cdots 2 \cdot 1} \\
\end{align}
here we could continue directly with
\begin{align}
f(n) = \frac{n^{n-1} n!}{(n-1)!} = n^{n-1} n = n^n
\end{align}
or complicate things with multiplying nominator and denominator with $n^{n-1}$
\begin{align}
f(n)
&= \frac{n^{2n-2}n!}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \frac{(n^2)^{n-1}n!}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \frac{n^2\cdot 2 n^2 \cdot 3n^2 \cdots (n-2) n^2 \cdot (n-1) n^2 \cdot n}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \frac{2 n^2 \cdot 3n^2 \cdots (n-2) n^2 \cdot (n-1) n^2 \cdot n^3}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \prod_{k=1}^{n-1} \frac{(k+1) n^2}{n^2-kn}
\end{align}
Steps $(2)$ to $(5)$
For even $n=2m+2$ and $m\ge 0$ one gets $2m+1$ factors:
\begin{align}
f(n)
=&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n}
\cdots
\frac{n^3-3n^2}{4n}
\cdot
\frac{n^3- 2n^2}{3 n}\cdot\frac{n^3- n^2}{2 n}\cdot n^2 \\
=&\underbrace{\frac{2 n^2}{n^2- n}\cdot
\underbrace{\frac{3 n^2}{n^2-2 n}\cdot
\underbrace{\frac{4 n^2}{n^2-3 n}
\cdots
\frac{n^2- 3n}{4}}
\cdot
\frac{n^2- 2n}{3}}\cdot\frac{n^2- n}{2}}_{\tiny \mbox{left factor} \times \mbox{right factor}}\cdot n^2 \\
=&(n^2)^m n^2 \\
=&n^{2m+2} \\
=&n^n
\end{align}
Note: The underbraces mean that $m$ pairs consisting of a left and a right side factor each, from the outside to the inside, multiply, resulting into $n^2$ each.
For odd $n=2m+3$ and $m\ge 0$ one gets $2m+1+1$ factors:
\begin{align}
f(n) 
&=(n^2)^m \cdot \frac{(m+2)n^2}{n^2-(m+1)n} \cdot n^2 \\
&=(n^2)^m \frac{m+2}{(2m+3)-(m+1)} n^3 \\
&=n^{2m+3} \\
&=n^n
\end{align}
So this seems to hold for $n\ge 2$.
