Why is $\lim(n!)^{1/n}=\infty$ 
Why is $\lim\limits_{n\to\infty}(n!)^{1/n}=\infty$

It is more or less clear that the sequence is increasing by ratio test $\frac{((n+1)!)^{1/(n+1)}}{(n!)^{1/n}}=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n+1]{(n)!}}\cdot\frac{\sqrt[n+1]{(n)!}}{\sqrt[n]{(n)!}}=\sqrt[n+1]{n+1}\cdot(n!)^{\frac{1}{n(n+1)}}$
but left multiplicand tends to $1$, what about the right one ? 
Is there convergence/divergence tests for sequences, I only found some for series ?
 A: It is obvious that in calculating n!, you are multiplying at least n/2 numbers which are greater than n/2. 
$(n!)^{1/n} > ((n/2)^{n/2})^{1/n} = (n/2)^{1/2}$ which is clearly divergent. 
A: According to Stirling's approximation
$$n!\sim\sqrt{2n\pi}(\frac ne)^n$$
so $$\sqrt[n]{n!}\sim(2n\pi)^{\frac{1}{2n}}(\frac ne)$$
since $$\lim_{n\to \infty}(2n\pi)^{\frac{1}{2n}}=1$$
$viz.$ $$\lim_{n\to\infty}(2n\pi)^{\frac2n}(\frac ne)=\infty$$
A: Use that
$$a_n>0\;\;and\;\;a_n\xrightarrow[n\to\infty]{}L\implies \sqrt[n]{a_1\cdot\ldots\cdot a_n}\xrightarrow[n\to\infty]{}L\;:$$
$$\frac1n\xrightarrow[n\to\infty]{}0\implies\sqrt[n]{1\cdot\frac12\cdot\ldots\cdot\frac1n}=\frac1{\sqrt[n]{n!}}\xrightarrow[n\to\infty]{}0$$
and we're done
A: Pick $n_0 \in \mathbb{N}$, then for $n \ge n_0$ we have
$n! \ge n_0^n {n_0! \over n_0^{n_0}}$ and so
$\sqrt[n]{n!} \ge n_0 \sqrt[n]{n_0! \over n_0^{n_0}}$.
Taking limits, we have $\liminf_n \sqrt[n]{n!} \ge n_0$. Since $n_0$ was
arbitrary we have the desired result.
A: Hint:
Show by induction that 
$$n!>\left(\frac{n}{e}\right)^n.$$ 
