I have to prove that $a_n$ is (strictly) increasing and diverges
$a_n = \sqrt[n]{n!}$ ; n $\in$ $\ \mathbb{N}$
From sequence I see that $a_n$ increasing to infinitive.
$\sqrt[1]{1!}=1 ,\ \sqrt[2]{2!} \approx 1.41, \ \sqrt[3]{3!} \approx 1.81 ,..., \sqrt[n]{n!} $
you want to see :
$\sqrt[n]{n!}< \sqrt[n+1]{(n+1)!}$
i.e., $n!<(n+1)!^{\frac{n}{n+1}}$
i.e., $n!^{n+1}< (n+1)!^n$
i.e., $n!*n!*\dots *n! \text{ (n+1 times)} < (n+1)!*(n+1)!,\dots * (n+1)! (\text{ n times})$
i.e., $n! < (n+1)(n+1)\dots (n+1) \text{ (n times)}$
i.e., $n.(n-1).(n-2).\dots 3. 2. 1 \text { (n terms)}< (n+1)(n+1)\dots (n+1) \text{ (n times)}$
you have $ i< (n+1) $ for all $i < n$
Atleat now yous should be able to see that this is true...
The sequence $a_n=\sqrt[n]{n}$ is increasing because $a_n\le a_{n+1}$ $$(n!)^{\frac{1}{n}}\le ((n+1)!)^{\frac{1}{n+1}}=((n+1))^{\frac{1}{n+1}}(n!)^{\frac{1}{n+1}}$$ that is $$ n!\le (n+1)^n $$ and this is true because it is simply $$ n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\le \underbrace{(n+1)(n+1)\cdots (n+1)}_{n \text{ times}} $$
Using Stirling's approximation $n!\sim \left(\frac{n}{e}\right)^n$, one has $$ \sqrt[n]{n!}\sim \frac{n}{e}\to \infty $$ so the sequence diverges.
$$\left(\frac{n}{2}\right)^{n/2}\leqslant n!\leqslant n^n$$
Sandwich theorem etc.
