Compute the limit of $\frac1{\sqrt{n}}\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)^{1/n^2}$ Compute the following limit:
$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}}  $$
I'm interested in almost any approaching way for this limit. Thanks.

Another way: by Stolz-Cesaro theorem, we have
$$\begin{align}
L&=\lim_{n\to\infty} \frac{1}{n^2} \left(\sum_{k=1}^{n} k\log (k) - \frac{n^2\log(n) }{2}\right)\\
&=\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}\\
&=\lim_{n\to\infty} \frac{1}{n^2-(n-1)^2} \left(n\log (n) - \frac{n^2\log(n) }{2}+\frac{(n-1)^2\log(n-1)}{2}\right)\\
&=\lim_{n\to\infty} \frac{1}{4n-2} \left(2n\log (n) -n^2\log(n)+(n-1)^2(\log(n)+\log(1-1/n))\right)\\
&=\lim_{n\to\infty} \frac{1}{4n-2} \left(-n+o(n)\right)=-\frac{1}{4}
\end{align}$$
and therefore
$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}}=e^{-1/4}.$$
