Solve $(\alpha,\beta)$ for $\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} = \beta$ Find the ordered pair $(\alpha,\beta)$ with non-infinite $\beta \ne 0$ such that $$\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} = \beta$$
My approach: 
$$\ln (1!2!\cdots n!) = (n)\ln 1 + (n-1)\ln 2 + \cdots + (2)\ln (n-1) + \ln(n) \\ \begin{align} = n\ln\left(\frac{1}{n}\right) + (n-1)\ln\left(\frac{2}{n}\right) + \cdots + \ln\left(\frac{n}{n}\right) + \frac{(n)(n+1)}{2} \ln (n)\end{align}$$
Then $$\ln(\sqrt[n^2]{1!2!\cdots n!}) = \frac{1}{n^2} \ln(1!2!\cdots n!) = \frac{n+1}{n} \cdot \ln n + \frac{1}{n} \left[\sum_{m=1}^n \left(\frac{n+1-m}{n} \cdot \ln \frac{m}{n}\right)\right]$$
And that's about as far as I got. Any ideas about proceeding with this method or perhaps even with a different method?
Thanks
A 

Another way: by Stolz-Cesaro theorem, we that
$$\begin{align}
L&=\lim_{n\to\infty} \frac{1}{n^{2}} \left(\sum_{k=1}^{n} \log(k!)-\alpha n^2\log(n) \right)\\
&=\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}\\
&=\lim_{n\to\infty} \frac{1}{n^2-(n-1)^2} \left(\log (n!) - \alpha n^2\log(n)+\alpha(n-1)^2\log(n-1)\right)\\
&=\lim_{n\to\infty} \frac{1}{2n-1} \left(\log (n!) -\alpha n^2\log(n)+\alpha(n-1)^2(\log(n)+\log(1-1/n))\right)\\
&=\lim_{n\to\infty} \frac{1}{2n-1} \left(n\log(n) -n+o(n) -2\alpha n\log(n)+\alpha(n-1)^2\Big(-\frac{1}{n}+o(1/n)\Big)\right)\\
&=\lim_{n\to\infty} \frac{1}{2n-1} \left((1-2\alpha)n\log(n) -(1+\alpha)n +o(n)\right)\\
&=\begin{cases}
\text{sgn}(1-2\alpha)\cdot +\infty &\text{if $\alpha\not=1/2$,}\\
-\frac{3}{4}&\text{if $\alpha=1/2$,}
\end{cases}
\end{align}$$
where we applied the Stirling approximation $\log (n !)=n\log n -n+o(n)$.
Therefore we may conclude that
$$\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} =e^L.$$
Hence $\alpha=1/2$ and $\beta=e^{-3/4}$.
