While verifying my answer to another question, I came across a problem of binomial coefficients:
Does $\hspace{.2cm}\displaystyle \prod_{k=1}^{n-1}\binom{n-1}{k}=\prod_{k=1}^{n-1}k^{2k-n}$ for all $n\in\mathbb{N}^+?$
I offer a proof by induction below, but I would appreciate references or alternative approaches.
On
The identity can be proved by induction on $n$. For $n=1$, each side is an empty product and so both equal $1$, so we assume it is true for some $n\geq 1$. Then for $n+1$ we have
$$\prod_{k=1}^{n}\binom{n}{k}=\prod_{k=1}^{n-1}\binom{n}{k}=\prod_{k=1}^{n-1}\frac{n}{n-k}\binom{n-1}{k}=\frac{n^{n-1}}{(n-1)!}\prod_{k=1}^{n-1}\binom{n-1}{k},\\ \prod_{k=1}^{n}k^{2k-n-1}=n^{n-1}\prod_{k=1}^{n-1}k^{2k-n-1}=\frac{n^{n-1}}{\prod_{k=1}^{n-1}k}\left(\prod_{k=1}^{n-1}k^{2k-n}\right)=\frac{n^{n-1}}{(n-1)!}\prod_{k=1}^{n-1}k^{2k-n}.$$
Then the inductive hypothesis implies that these expressions are equal, and we conclude that the identity is true for all $n\in \mathbb{N}^+$.
Well, in each term of your product, the numerator is $(n-1)!$. Thus, the "total" numerator is $$\left((n-1)!\right)^{n-1}=\prod_{1}^{n-1}k^{n-1}$$
As for the denominator, each individual denominator is a product of 2 factorials, the argument of first factorial "grows" from $0$ to $(n-1)$, the argument of the second one "grows smaller" from $n-1$ to $0$. Thus, each argument appears exactly two times, and we have $\left(\prod_1^{n-1}k!\right)^2$. Note that in $\left(\prod_1^{n-1}k!\right)$ we come across number $k$ $n-k$ times, thus our denominator is $$\left(\prod_1^{n-1}k!\right)^2 = \prod_1^{n-1}k^{2(n-k)}$$
Trivial division allows to conclude. I favour this kind of proofs because you make virtually no formula manipulation, almost impossible to get it wrong.