Closed form of $\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$ I was wondering if there is a closed form for
$$\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$$
I know that for 
$$\prod_{i=0}^{N}\big(i!\big)=G(N+2)$$
where we have expressed it as Barnes G-function. Yet I am not sure how to go about this with the binomial involved? 
 A: An asymptotic formula can be derived that is better than 0.1% accurate for $n>7.$  I'll work with the logarithm and sketch a proof that
$$ (*) \quad S_n :=\sum_{k=0}^n \binom{n}{k} \log \Gamma(k+1) \sim 
2^n\Big( \log \Gamma(n/2+1) + 1/4 - 1/(8n) + ... \Big) $$
Use the following formula (it can be found on the wiki for $\Gamma$ )
$$ \log \Gamma(k+1) = -\gamma \, k + \int_0^\infty \frac{e^{-k\,t} -1 + k \,t}{e^t-1} \frac{dt}{t} $$
Insert the previous equation into the left-hand side of (*), switch the finite sum with the integral, do the sum, and factor out a $2^n,$
$$ S_n = 2^n\Big(-\gamma \frac{n}{2} + \int_0^\infty \frac{e^{-n\,t/2}\cosh^n(t/2) -1 + \frac{n}{2} \,t}{e^t-1} \frac{dt}{t} \Big)$$
Note I've used a hyperbolic trig ID.  The reason is because, for $t$ small enough,
$$ \cosh^n(t/2) = 1 + \frac{n\,t^2}{8} + \frac{(3n-2)n\,t^4}{384}+...$$
Therefore, 
$$ S_n \sim 2^n\Big( -\gamma \frac{n}{2} +  \int_0^\infty \frac{e^{-n\,t /2} -1 + \frac{n}{2} \,t}{e^t-1} \frac{dt}{t}
+\int_0^\infty \frac{dt}{t} \frac{e^{-n\,t/2 }}{e^t-1}
\big(\frac{n\,t^2}{8} + \frac{(3n-2)n\,t^4}{384} \big) \Big)$$
$$ = 2^n\Big(\log\Gamma(1+n/2)+\frac{n}{8}\int_0^\infty \frac{e^{-n\,t/2 }}{e^t-1}t\,dt + \frac{(3n-2)n}{384}\int_0^\infty \frac{e^{-n\,t/2 }}{e^t-1}t^3\,dt + ... \Big)$$
We can split the integral up because each converges (the $t$ is denominator is cancelled).  More importantly the terms shown, and the subsequent ones, form an asymptotic sequence in $n$.  In particular, the integrals can be explicitly written in terms of odd derivatives of the polygamma function, and since the polygamma function asymptotics is well-known, you can derive (by hand, if you want) the terms given in (*).  
The remark about the accuracy of the expansion is from some numerical calculations.  Take the $LHS(*)/RHS(*)$ and for n=7 the ratio is 0.99949 and for n=20, the ratio is 0.999993
Incidentally, the asymptotics proves that the product form will not be a Barnes G-function.
