Proving $\big(n!^{\frac1n}\big)_{n\in\mathbb N^*} \to \infty$ By definition, for any $a\in\mathbb R$, there exists $k\in\mathbb N^*$ such that, if $n\in\mathbb N+k$, then $n!^{\large\frac1n}>a$. Therefore, by induction, I must:


*

*find some $k$ that satisfies $k!>a^k$;

*show that, for each $n\ge k$, if $n!>a^n$, then $(n+1)!>a^{n+1}$.


For the 2nd part, it's enough that $k\ge a-1$. I noticed the smallest adequate $k$ is $4$ for $a=2$ and $7$ for $a=3$, and I'm convinced that one can always find a $k$, but I would like a function of $a$ to make an actual argument.
Definition aside, methods other than this one are welcome. I tried picking a sequence dominated by $\big(n!^{\large\frac1n}\big)$ that went to infinity as well, but couldn't find any.

Using Stirling's approximation, we have $n!> \sqrt{2\pi n}(n/e)^n$.  Then $(n!)^{1/n}> \sqrt[n]{2\pi}n^{1/n} \frac{n}{e}$.  Since $\sqrt[n]{2\pi}\rightarrow 1$ and $n^{1/n}\rightarrow 1$ but $\frac{n}{e}\rightarrow \infty$, we see that $(n!)^{1/n}\rightarrow \infty$.
