I've been trying to prove:
$$ \nu_{{n}} \left(\prod^{n-1}_{j=1} {n\choose j} \right)=n-1 \quad \operatorname{if and only if} \quad n \in \mathbb P\cup{\{1}\}$$
From the basic properties of the p-adic order of a number:
$$\nu_{{p}} \left( {\frac {n!}{j!\, \left( n-j \right) !}} \right) =\nu_ {{p}} \left( n! \right) -(\nu_{{p}} \left( j! \right) +\nu_{{p}} \left( \left( n-j \right) ! \right)) $$
Therefore:
$$\nu_{{p}} \left(\prod^{n-1}_{j=1} {\frac {n!}{j!\, \left( n-j \right) !}} \right) =\sum^{n-1}_{j=1}\Bigl(\nu_ {{p}} \left( n! \right) -(\nu_{{p}} \left( j! \right) +\nu_{{p}} \left( \left( n-j \right) ! \right))\Bigr)=(n-1)\nu_ {{p}} \left( n! \right)-\sum^{n-1}_{j=1}(\nu_{{p}} \left( j! \right) +\nu_{{p}} \left( \left( n-j \right) ! \right)) $$
Allowing us to restate the original lemma as:
$$(n-1)\nu_ {{n}} \left( n! \right)-\sum^{n-1}_{j=1}(\nu_{{n}} \left( j! \right) +\nu_{{n}} \left( \left( n-j \right) ! \right))=n-1 \quad \operatorname{if and only if} \quad n \in \mathbb P\cup{\{1}\}$$
But I don't feel as if I have gotten anywhere here.
I then wanted to just say the denominator of $$\frac{\prod^{n-1}_{j=1} {n\choose j}}{n^n}$$
is $n$ if and only if $n$ is prime or $n=1$, but that's just handballing the absence of proof to another statement really.