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Evaluate the following limit: $$\lim_{n\rightarrow +\infty}\frac{\left ( 1^1\cdot 2^2\cdot 3^3\cdots n^n \right )^{1/n}}{\sqrt{n}}$$ plus the original limit which is: $$\lim_{n\rightarrow +\infty}\frac{\left ( 1^1\cdot 2^2\cdot 3^3\cdots n^n \right )^{\color{red}{1/n^2}}}{\sqrt{n}}$$

For every \(n\in\mathbb{N}\) we have \begin{align*}
\textstyle\prod_{k=1}^{n}k^k>n^n\quad&\Rightarrow\quad\Bigl({\textstyle\prod_{k=1}^{n}k^k}\Bigr)^{\frac{1}{n}}>\bigl({n^n}\bigr)^{\frac{1}{n}}=n\\
&\Rightarrow\quad\frac{\Bigl({\textstyle\prod_{k=1}^{n}k^k}\Bigr)^{\frac{1}{n}}}{\sqrt{n}}>\frac{n}{\sqrt{n}}=\sqrt{n}
\end{align*} Because \(\lim_{n\rightarrow +\infty}\sqrt{n}=+\infty\) we have that \[\lim_{n\rightarrow +\infty}\frac{\Bigl({\textstyle\prod_{k=1}^{n}k^k}\Bigr)^{\frac{1}{n}}}{\sqrt{n}}=+\infty\,.\]