I could not understand the accepted answer. But I tried to solve it by constructing an inequality, whose one side is easy to compute and that computation is enough to conclude on the other side.
For any odd natural number $n$,
\begin{align*}
n!\,n!&=(1\cdot 2 \cdots n)\times (1\cdot 2 \cdots n)\\
&=(1\cdot n)^2\times \{2\cdot(n-1)\}^2 \times \cdots \times \{(\frac{n+1}{2})\cdot (\frac{n+1}{2})\}^2\\
&> \Big(\frac{n}{2}\Big)^2 \times \Big(\frac{n}{2}\Big)^2 \times \cdots \Big(\frac{n}{2}\Big)^2 \,\,\Big[\Big(\frac{n+1}{2}\Big) \,\,\text{times}\Big]\\
&=\Big(\frac{n}{2}\Big)^{2\big(\frac{n+1}{2}\big)}=\Big(\frac{n}{2}\Big)^{n+1}
\end{align*}
Then $(n!)^{\frac{2}{n}}>\big(\frac{n}{2}\big)^{\frac{n+1}{n}} \implies (n!)^{\frac{1}{n}}>\big(\frac{n}{2}\big)^{\frac{n+1}{2n}}=\big(\frac{n}{2}\big)^{\frac{1}{2}+\frac{1}{2n}}$
Similarly, for any even natural number $n$,
$$n!\,n!>\Big(\frac{n}{2}\Big)^n \,\,\Big[\text{ There will be } \frac{n}{2} \text{ terms}\Big]$$
Then $(n!)^{\frac{2}{n}}>\big(\frac{n}{2}\big) \implies (n!)^{\frac{1}{n}}>\big(\frac{n}{2}\big)^{\frac{1}{2}}$
Using the fact that $\lim_{n \to \infty} n^{\frac{1}{n}}=1$, one can obtain $\lim_{n \to \infty} \big(\frac{n}{2}\big)^{\frac{1}{2n}}=1$
Thus in all cases we have $\lim_{n \to \infty}(n!)^{\frac{1}{n}}\geq\lim_{n \to \infty}\big(\frac{n}{2}\big)^{\frac{1}{2}}$
The limit in the R.H.S. diverges. Hence the desired limit (in the L.H.S.) also goes to $\infty$.