TITLE: Find $\displaystyle{\lim_{n \to \infty}} \sqrt[n]{n!}$ without Stirling's approximation.
QUESTION [5 upvotes]: Okay I've seen many posts as this one, but in every post it uses Stirling(I can't use it because I've never learnt it), or weird equalities with logs (I also never learnt this) or other methods I'm not allowed to use in my exam.
I have to prove that $\lim n!^{(1/n)}$ = $\infty$, only by using D'Alembert, $n^{th}$ root test, Stolz, squeeze theorem or by playing with inequalities etc 
What I thought is that I could prove that $\lim \frac{1}{n!^{(1/n)}} = 0$ (again only using those basic theorems) and that would mean that $\lim n!^{(1/n)} = \infty$ but I'm not sure if that's correct.
Please don't mark as duplicate! I need an answer and I haven't found any post with an answer I can use!!

REPLY [4 votes]: Note that since $n!\ge(n/2)^{n/2}$, we have 
$$\sqrt[n]{n!}\ge \sqrt{n/2}$$