Proof:
Since we have $n!\geq \left(n/2\right)^{n/2} \to \infty$, we can conclude that $\lim\limits_{n\to\infty}\sqrt[n]{n!}=\infty$. I understand that by taking the n-th root we have $\sqrt{n!}\geq n/2$. But why is the inequality true? And is it enough to be "equal or greater than" if solely for $n=0$ the expression $0!=(0/2)^{0/2}$ match?
I will show by completely elementary means that
$1+1/m \lt \dfrac{(m!)^{1/m}}{m/e} \lt (1+3/\sqrt{m})(1+1/m^2)(1+1/m) $.
I will first show that if $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$ then $a_n < a_{n+1}$ and $b_n > b_{n+1}$.
Since $a_n < b_n$ and $b_n-a_n =\frac1{n}(1+1/n)^n \to 0 $, this shows that $\lim_{n \to \infty} a_n$ and $\lim_{n \to \infty} b_n$ exist and are equal.
We use the very ingenious proof in N.S Mendelsohn, "An application of a famous inequality", Amer. Math. Monthly 58 (1951), 563, which uses the AGMI, which we will use in the form $((v_1+v_2+...v_n)/n)^n > v_1v_2...v_n$ (all $v_i$ positive) with equality if and only if all the $v_i$ are equal (this allows us to avoid the use of n-th roots).
For $a_n$, consider $n$ values of $1+1/n$ and 1 value of 1. By the AGMI, $((n+2)/(n+1))^{n+1} > (1+1/n)^n $, or $(1+1/(n+1))^{n+1} > (1+1/n)^n $, or $a_{n+1} > a_n $.
For $b_n$, consider $n$ values of $1-1/n$ and 1 value of 1. By the AGMI, $(n/(n+1))^{n+1} > (1-1/n)^n$ or $(1+1/n)^{n+1} < (1+1/(n-1))^n$ , or $b_n < b_{n+1}$.
Since the $a_n$ are increasing, the $b_n$ are decreasing, and $a_n < b_n$, all the $a_n$ are less than any of the $b_n$. Since $b_5 = 2.9859... < 3 $, all the $a_n$ are less than 3.
If n > m, since $a_n < b_m$ and $b_m-a_m = a_m/m $, $a_n-a_m < b_m-a_m = a_m/m < 3/m $.
The common limit is called $e$.
Since $a_n < e < b_n$ for all $n$,
$\begin{array}\\ e^m &\gt \prod_{n=1}^m a_n\\ &= \prod_{n=1}^m (1+1/n)^n\\ &= \prod_{n=1}^m \dfrac{(n+1)^n}{n^n}\\ &= \dfrac{\prod_{n=1}^m (n+1)^n}{\prod_{n=1}^m n^n}\\ &= \dfrac{\prod_{n=2}^{m+1} n^{n-1}}{\prod_{n=1}^m n^n}\\ &= \dfrac{\prod_{n=2}^{m+1} n^{n-1}}{m!\prod_{n=1}^m n^{n-1}}\\ &= \dfrac{(m+1)^m}{m!}\\ \text{so}\\ m! &\gt \dfrac{(m+1)^m}{e^m}\\ \text{or}\\ (m!)^{1/m} &\gt \dfrac{m+1}{e}\\ \text{or}\\ \dfrac{(m!)^{1/m}}{m/e} &\gt 1+1/m\\ \end{array} $
Similarly,
$\begin{array}\\ e^m &\lt \prod_{n=1}^m b_n\\ &= \prod_{n=1}^m (1+1/n)^{n+1}\\ &= \prod_{n=1}^m \dfrac{(n+1)^{n+1}}{n^{n+1}}\\ &= \dfrac{\prod_{n=1}^m (n+1)^{n+1}}{\prod_{n=1}^m n^{n+1}}\\ &= \dfrac{\prod_{n=2}^{m+1} n^{n}}{\prod_{n=1}^m n^{n+1}}\\ &= \dfrac{\prod_{n=2}^{m+1} n^{n}}{m!\prod_{n=1}^m n^{n}}\\ &= \dfrac{(m+1)^{m+1}}{m!}\\ \text{so}\\ m! &\lt \dfrac{(m+1)^{m+1}}{e^m}\\ &= \dfrac{(m+1)(m+1)^{m}}{e^m}\\ \text{or}\\ (m!)^{1/m} &\lt \dfrac{(m+1)^{1/m}(m+1)}{e}\\ \text{or}\\ \dfrac{(m!)^{1/m}}{m/e} &\lt (m+1)^{1/m}(1+1/m)\\ &= m^{1/m}(1+1/m)^{1/m}(1+1/m)\\ \end{array} $
If $(1+1/m)^{1/m} = 1+h_m$, $1+1/m =(1+h_m)^m \gt 1+mh_m $ or $h_m < 1/m^2$ so $(1+1/m)^{1/m} \lt 1+1/m^2 $.
Since $(1+m^{-1/2})^m \gt 1+m^{1/2} \gt m^{1/2} $, raising to the $2/m$ power, $m^{1/m} \lt (1+m^{-1/2})^2 =1+2m^{-1/2}+m^{-1} \le 1+3m^{-1/2} $.
Therefore $\dfrac{(m!)^{1/m}}{m/e} \lt (1+3m^{-1/2})(1+1/m^2)(1+1/m) $.
note that from Sterling's approximation: $$n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n$$ and so: $$n(n!)^{-1/n}\sim n(2\pi n)^{-1/2n}\frac{e}{n}=e(2\pi n)^{-1/2n}$$ and we can now say: $$\lim_{n\to\infty}e(2\pi n)^{-1/2n}=\lim_{n\to\infty}\frac{e}{(2\pi)^{1/n}}=e$$
proof-explanationmeans you want explanation for some established proof. Are you actually asking how to prove the limit in the title? If yes, then see this: https://math.stackexchange.com/q/201906/9464 – Jun 18 '19 at 22:28