Prove that $\sqrt[n]{n!}$ is increasing and diverges 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!} $
 A: 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... 
A: 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.
A: $$\left(\frac{n}{2}\right)^{n/2}\leqslant n!\leqslant n^n$$
Sandwich theorem etc.
