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.

You have to prove: 
$$ \forall a > 0 \, \exists n \, \ni \, n! > a^n $$
which is equivalent to
$$ \forall a > 0 \, \exists n \, \ni \, \frac{a^n}{n!} < 1 \, .
$$
This last statement follows from the fact that
$$
e^a = \sum_{n = 0}^\infty \frac {a^n}{n!}
$$
is convergent for all $ a > 0$. 
