How to rigorously find $\lim_{n \to \infty} \sqrt[n]{n!}$? Let $x_n \colon= \sqrt[n]{n!} $
for all $n \in \mathbb{N}$. 
Then how to determine rigorously whether the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converges or diverges? 
And, how to find $\lim_{n \to \infty} x_n$, rigorously?
By rigorously I mean using the same machinery as has been developed by Rudin until Chap. 3, where he discusses sequences. 
My Attempt: 

We note that, for all $n \in \mathbb{N}$ such that $n > 1$, we have
  $$ 1 \leq \left( x_n \right)^n \leq n^{n-1}, $$
  and so 
  $$ 1 \leq x_n \leq n^{ (n-1)/n } = \frac{ n }{\sqrt[n]{n} }. $$
  However, although 
  $$ \lim_{n \to \infty} \sqrt[n]{n} = 1, $$
  we also have 
  $$ \lim_{n \to \infty} n = +\infty. $$
  Thus the squeeze theorem is not applicable. 

Or, can we find some majorizing sequence converging to $1$?
Is this sequence monotonic? 
 A: Hint: Consider the series $\sum \dfrac{1}{n!}$ and use the root test.
A: You can derive it form Stirling's formula:$$\sqrt{2\pi}n^{n+\frac12}e^{-n}\leqslant n!\leqslant en^{n+\frac12}e^{-n}.$$
A: By ratio-root criteria 
$$a_n = \sqrt[n]{n!} \quad b_n=n!$$
$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$
thus since
$$\frac{b_{n+1}}{b_n}=\frac{{(n+1)!}}{n!}=n+1\to+\infty \implies a_n = \sqrt[n]{n!}\to +\infty $$
A: Taking log plus Cesàro-Stolz:
$$
\lim_{n\to\infty}\log\sqrt[n]{n!} =
\lim_{n\to\infty}\frac{\log 1 + \cdots + \log n}n =
\lim_{n\to\infty}\log n = \infty,
$$
so
$$
\lim_{n\to\infty}\sqrt[n]{n!} = \infty.
$$
A: From 
$$(2n-k)(n-k)\gt(2n-2k)(n-k)=2(n-k)^2$$ 
for $0\lt k\lt n$, we have
$$\begin{align}
x_{2n}
&=\sqrt[2n]{(2n)!}\\&=\sqrt[2n]{(2n)(n)(2n-1)(n-1)\cdots(n+2)(2)(n+1)(1)}\\
&\gt\sqrt[2n]{2(n)^22(n-1)^2\cdots2(2)^22(1)^2}\\
&=\sqrt2\sqrt[n]{n!}
\end{align}$$
for $n\ge2$. (The inequality is not strict for $n=1$, because there are no $k$'s between $0$ and $1$.) Thus the sequence has subsequences, at least, that diverge to infinity, for example
$$x_{2^{n+1}}\gt(\sqrt2)^n\to\infty$$
To show that the sequence as a whole diverges to infinity, it suffices to show that $x_{n+1}\gt x_n$, i.e.,
$$((n+1)!)^n\gt(n!)^{n+1}$$
which is to say,
$$(n+1)^n\gt n!$$
which is obviously true.
A: By elementary means:
Consider the sequence 
$$2^0,2^1,2^1,2^2,2^2,2^2,2^2,2^3,2^3,2^3,2^3,2^3,2^3,2^3,2^3,\cdots2^b,2^b,2^b,\cdots 2^b$$
term-wise inferior to the naturals.
We obviously have 
$$n!\ge 2^{b2^b}=2^{b(n+1)/2}$$
hence the series diverges.
