Prove that $n!^{\frac{1}{n}}$ is unbounded

Question

I would like to show that $n!^{\frac{1}{n}}$ is unbounded. I cannot seem to find a $k(M)$ s.t. $k!^{\frac{1}{k}}>M$ for any $M\in \mathbb{R}$.

Answer 1

We have $$ n! = n \cdots (n/2) \cdots 1 \ge (n/2)^{n/2} $$ Therefore, $$ ({n!})^{1/n} \ge n^{1/2} = \sqrt{n}\to \infty $$

Answer 2

Consider $$A_n=(n!)^{\frac{1}{n}}\implies \log(A_n)={\frac{1}{n}}\log(n!)$$ Now, use Stirling approximation $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ which makes $$\log(A_n)=\log (n)-1+O\left(\frac{1}{n}\right)\implies A_n=e^{\log(A_n)} \sim \frac n e+\frac{\log(2\pi n)}{2e}$$

For $n=10$, the "exact" value is $\approx 4.52873$ while the above approximation gives $\approx 4.44039$.