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 ?

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 $$\lim_{n\to \infty}(2n\pi)^{\frac{1}{2n}}=1$$
$viz.$ $$\lim_{n\to\infty}(2n\pi)^{\frac2n}(\frac ne)=\infty$$
