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.

Here is a simple but not-so-elegant proof.
Without loss, it is enough to prove the result for fixed natural a > 1. Showing that $(n!)^{1/n} > a$ eventually(ev.) is the same as showing 
$$n! > a^n  \quad\text{ev.} \iff x_n := \frac{n!}{a^n} > 1\quad\text{ev.}$$ 
For $n>a$, 
$$x_n = \frac{a!}{a^a} \frac{n(n-1)…(a+1)}{a^{n-a}} $$
Every term on the top of the second fraction is at least a+1, and there are $n-(a+1)+1 = n-a$ terms, so 
$$ x_n \geq \frac{a!}{a^a} \frac{(a+1)^{n-a}}{a^{n-a}} = \frac{a!}{a^a} \left( \frac{a+1}{a} \right)^{n-a} = C(a)  \left( \frac{a+1}{a} \right)^{n-a} $$
($C(a) = \frac{a!}{a^a} > 0$ is constant in $n$) which is bigger than 1 eventually, hence the result.
