I will slightly expand upon my comment. For N$N$ a prime, one has the desired product being (N-1)!$(N-1)!$. For N=pq$N=pq$, a product of two primes, the desired product is N! * pq/[p!q^p q!p^q]$\frac{N! pq}{p!q^p q!p^q}$ . I have not worked it out, but one should note a similarity between this representation and that of the cyclotomic polynomial Phi_pq(x)$\Phi_pq(x)$. For square free N=pqr$N=pqr\ldots$ I imagine an analogous relation holds.
EDIT 2012.05.19 Thanks to Gjergji Zaimi, the above idea can be extended and nicely expressed in terms of the Moebius function $\mu()$ and the Euler totient function $\phi()$ as $$N^{\phi(N)}\prod_{d|N}\left(\frac{d!}{d^d}\right)^{\mu(N/d)} .$$ (I have faith enough to post it, but have not fully verified it myself. It looks right to me.) END EDIT 2012.05.19
If you want something rougher but still in the ball park, consider (N!)^2$(N!)^2$ as product of terms of the form (n+1-i)i$(n+1-i)i$, which for many i look like o(N^2)$o(N^2)$. Then the desired product is is something like phi(n)/n$\phi(N)/N$ fraction of those terms, perhaps with a power of e to take care of the discrepancy. So I nominate without proof [((N!)^(1/N))/e]^phi(N)$[(N!)^{1/N}/e]^{\phi(N)}$ as a rough estimate estimate. I imagine provable upper and lower bounds are a mere exponential factor away from this estimate estimate.
Gerhard "Ask Me About Rough Estimates" Paseman, 2012.05.18