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.
 A: When $n$ is even, $n/2$ of the numbers in the sequence $1,\ldots,n$ are greater than $n/2$. Therefore, $n! \geq (n/2)^{n/2}$, and this leads to the desired result.
A: Start with the logarithm of what we want
$$
\begin{align}
\frac1n\log(n!)
&=\frac1n\sum_{k=1}^n\log(k)\\
&\ge\frac1n\int_1^n\log(x)\,\mathrm{d}x\\
&=\log(n)-1+\frac1n
\end{align}
$$
Therefore,
$$
\left(n!\right)^{1/n}\ge\frac ne
$$
A: Using AM-GM:
$$\frac{\frac{1}{1} + \frac{1}{2} +\cdots + \frac{1}{n}}{n} \ge \sqrt[n]{\frac{1}{n!}}$$
$$\sqrt[n]{n!} \ge \frac{n}{\frac{1}{1} + \frac{1}{2} +\cdots + \frac{1}{n}} = \frac{n}{H_n}$$
But since $H_n$ grows as $\ln x$ , we have (as $n\to\infty$):
$$\frac{n}{\frac{1}{1} + \frac{1}{2} +\cdots + \frac{1}{n}}  \to \infty$$
and therefore (as $n\to\infty$),
$$\sqrt[n]{n!} \to \infty$$
A: 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 $(n!)^{1/n}\rightarrow \infty$.
A: 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.
A: 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$. 
