I need help showing that $$\lim_{n \to \infty} \frac{((n-1)!)! (n-1)!^{n!}}{(n!)!} = 0.$$ If I take logarithm, I get $\begin{align*} \lim_{n \to \infty} \ln(\frac{((n-1)!)! (n-1)!^{n!}}{(n!)!}) &= \lim_{n \to \infty} \ln(((n-1)!)!) + n! \ln((n-1)!) - \ln((n!)!) \\ &= \lim_{n \to \infty} \ln(n-1)+\ln(n-3)+...+\ln(3)\\& + n! (\ln(n-1)+\ln(n-2)+...+\ln(2))\\& - \ln(n) - \ln(n-2) - ... - \ln(2), \end{align*}$
but I don't see if that even helps.
Let, $m=(n-1)!$, Then $n!=n.(n-1)!=mn$, $m>>n$
Therefore you have $$L=\lim_{n\to\infty} \frac{m!m^{mn}}{(mn)!}$$
If we take logarithm and use Stirling's formula for large $k$, $ln(k!)=k.ln(k)-k$, we would have, $$ln(L)=\lim_{m>>n, n\to\infty} m.ln(m)-m+mn.ln(m)-mn.ln(mn)+mn$$ Now recall, $mn=n!$, and $ln(m)=ln((n-1)!)=(n-1)ln(n-1)-(n-1)$, and show with patient calculation that $$ln(L)=\lim_{m>>n, n\to\infty}=-\infty$$ Thus $L=0$
Another approach is to show that, for $n>1$, $$ \frac{m!m^{mn}}{(mn)!} < \frac{1}{n}$$ You may use induction method for that.