Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!...n!)^2} \right)$ 
Evaluating $$\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!...n!)^2} \right)$$

I'm not quite sure where to start in evaluating this. Some pointers, or a solution, would greatly be appreciated.
 A: Since
$$
\int_1^n\log(x)\,\mathrm{d}x\le\log(n!)\le\int_1^{n+1}\log(x)\,\mathrm{d}x\tag{1}
$$
we have that
$$
\log(n!)=n\log(n)-n+O(\log(n))\tag{2}
$$
therefore,
$$
\log\left(\frac{n^{n-1}}{n!}\right)=n+O(\log(n))\tag{3}
$$

Note that
$$
\left.\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\middle/\frac{((n-1)!)^{n-1}}{(0!1!2!\cdots(n-1)!)^2}\right.
=\frac{n^{n-1}}{n!}\tag{4}
$$
Applying $(3)$, we have
$$
\begin{align}
\log\left(\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\right)
&=\sum_{k=1}^n\log\left(\frac{k^{k-1}}{k!}\right)\\
&=\sum_{k=1}^n(k+O(\log(k)))\\
&=\frac{n^2}2+O(n\log(n))\tag{5}
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}\frac1{n^2}\log\left(\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\right)=\frac12\tag{6}
$$
A: No need to reference Barnes G-functions.  These are integers after all.
$$\log{\left ( \frac{(n!)^n}{(0!1!2!...n!)^2} \right )} = n \log{n!} - 2 \sum_{k=0}^n \log{k!}$$
Now
$$\sum_{k=0}^n \log{k!} = \sum_{k=0}^n \log{\Gamma(k+1)} = \sum_{k=1}^{n+1} \log{\Gamma(k)}$$
Now use Euler-Maclurin:
$$\begin{align}\sum_{k=1}^{n+1} \log{\Gamma(k)}&= \int_0^{n+1} dx \, \log{\Gamma(x)} + \frac12 \left [\log{\Gamma(n+1)} - \log{\Gamma(1)} \right ] + o(n)\end{align}$$
Now, we use Raabe's integral, which is
$$ \int_{k}^{k+1} dx \, \log \Gamma(x) = \frac12 \log{(2 \pi)} + k \log{k}-k$$
so that
$$\begin{align} \int_{0}^{n+1} dx \, \log{\Gamma(x)} &= \sum_{k=0}^n \int_{k}^{k+1} dx \, \log{\Gamma(x)}\\ &= \frac{n+1}{2}  \log (2 \pi) - \frac{n (n+1)}{2} +  \sum_{k=1}^n k \log{k} \\ &= \frac{n+1}{2}  \log (2 \pi) - \frac{n (n+1)}{2} + \log{(1^1 2^2 \cdots n^n)} \\ &= \frac{n+1}{2}  \log (2 \pi) - \frac{n (n+1)}{2} + \log{\left [\frac{n!^n}{(n-1)!(n-2)!^2 \cdots 2^{n-2} 1^{n-1}} \right ]} \\ &= \frac{n+1}{2}  \log (2 \pi) - \frac{n (n+1)}{2} + n \log{n!} - \sum_{k=1}^{n-1} \log{k!}\end{align}$$
Putting this all together, we find that
$$2 \sum_{k=1}^{n} \log{k!} =  \frac{n+1}{2}  \log (2 \pi) - \frac{n (n+1)}{2} + n \log{n!} + \frac{3}{2} \log{n!} +o(n) $$
so that, finally, using the fact that Stirling's approximation is equivalent to $\log{n!} = n \log{n} - n + o(n) $,
$$\lim_{n \to \infty} \frac1{n^2} \left ( n \log{n!} - 2 \sum_{k=1}^{n} \log{k!} \right ) = \frac12 $$
ADDENDUM
Raabe's integral is actually not all that hard to evaluate.  Rewrite as
$$\int_k^{k+1} dx \, \log{\Gamma(x)} = \int_0^1 dx \, \log{\Gamma(x+k)} $$
$$\log{\Gamma(x+k)} = \log{\Gamma(x)} + \sum_{m=0}^{k-1} \log{(x+m)} $$
Thus,
$$\begin{align}\int_k^{k+1} dx \, \log{\Gamma(x)} &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} \int_0^1 dx \, \log{(x+m)}\\ &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} [(m+1) \log{(m+1)} - m \log{m} - (m+1)+m] \\ &= \int_0^1 dx \, \log{\Gamma(x)} + k \log{k} - k \end{align}$$
To evaluate the integral on the RHS, use the duplication formula:
$$\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right ) = 2^{1-x} \sqrt{\pi} \Gamma(x) $$
so that
$$\log{\Gamma(x)} = \log{\left [\frac{\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right )}{\sqrt{\pi} 2^{1-x}} \right ]} = -\frac12 \log{\pi} - \log{2} + \log{\Gamma \left ( \frac{x}{2} \right )} + \log{\Gamma \left ( \frac{x+1}{2} \right )} + x \log{2}$$
Thus,
$$\begin{align}\int_0^1 dx \, \log{\Gamma(x)} &= -\frac12 \log{(2 \pi)} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x}{2} \right )}  + \int_0^1 dx \, \log{\Gamma \left ( \frac{x+1}{2} \right )}\\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1/2} dx \, \log{\Gamma(x)} + 2 \int_{1/2}^1 dx \, \log{\Gamma(x)} \\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1} dx \, \log{\Gamma(x)} \end{align}$$
The result follows.
A: 1). Stirling's approximation and Stirling-like asymptotic series for Barnes G- function here at $(14)$ http://mathworld.wolfram.com/BarnesG-Function.html
Therefore, 
$$\lim_{n\rightarrow \infty} \dfrac{\displaystyle \ln \left ( \frac{(n!)^n}{(0!1!2!…n!)^2} \right )}{n^2}=\frac{1}{2}.$$
2). Apply Stolz–Cesàro theorem for an elementary solution.
3). Combine the squeeze theorem, Stirling's approximation and the Riemann sums.
4). Make use of the Euler–Maclaurin formula.
