Proving that $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$ I have to prove this
$$\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$$
My proof is the following:
By using $$n!>n^{\frac{n}{2}}$$
$$\frac{1}{n!}<\frac{1}{n^{\frac{n}{2}}}$$
$$\left(\frac{1}{n!}\right)^{\frac{1}{n}}<\left( \frac{1}{n}\right)^{\frac{n}{2}\cdot\frac{1}{n}}=\frac{1}{\sqrt{n}}\underset{n\to \infty}{\to}0$$
Using the Squeeze theorem
$$0<\frac{1}{\sqrt[n]{n!}}<\frac{1}{\sqrt{n}} \implies \frac{1}{\sqrt[n]{n}}\underset{n\to \infty}{\to} 0$$
I think I have done something wrong. Can I even use the first inequality here? and if yes-do I have to prove that too? and is it Ok that I write too little words in my proofs?. Could you please help me?
 A: To start of, the first inequality is not correct for all possibilities of n, say for example $n = 2$, then $2! = 2$ and $2^{\frac{2}{2}} = 2$, the strict inequality also doesn't work for $n=1$. So either use a non strict inequality or define it only for $\forall n \in \mathbb{N}: n > 2 $. If you propagate the non strict inequality in the rest of your proof then the steps you take are as far as I know correct.
Secondly, if you want to use the first inequality you will have to prove it, by proving it you would have noticed that it doesn't work for these smaller values. You can try proving the first inequality by thinking about what both sides of the inequality look like when written in full. I won't spoil the proof for that, but if you have trouble finding it you can always ask follow up questions in the comments.
To answer the second question it depends, if you are just writing the proof for yourself it would be sufficient to just write this to check if you're proof works. But if you're writing this for an assignment or an exam or a paper or anything else than a scribble on paper more text with clarification of what you're doing would always be better to write a clear proof. If a proof is correct in all steps that doesn't make it a good proof, a good proof should be as clear as possible for the reader. Also the third expression could also be split for a clearer proof in the form of first showing the inequality then showing the limit and reffering to those in your Squeeze theorem. E.g. using the found inequality we apply the Squeeze theorem wich results in...
Good luck!
A: Another answer has covered how much descriptive detail to include in your proof.  The objective is to ensure your argument is clear, not just to yourself but to another reader who may not be able to follow each step without some help.
The crucial inequality is the first $n! > n^{n/2}$ for sufficiently large $n$.  In fact, for your argument you do need a strict inequality;  your subsequent reasoning would work just as well if $n! \geqslant n^{n/2}$ for $n$ greater than some threshold.  To establish it you can take (natural) logarithms. $n! = \log n+\log(n-1)+\cdots+\log 1$.  Now consider a simple step function,
$$ f(x) = \left\{\array{0 & 0 < x \leqslant 1 \\ \log 2 & 1 < x \leqslant 2 \\ \cdots &\cdots \\ \log n & n-1 < x \leqslant n \\ \cdots & \cdots} \right.$$
This is constructed deliberately so that for $x \geqslant 1$, $f(x)\geqslant \log x$.  Then,
\begin{align}
\log n! &= \int_1^n f(x) ~dx \\ & \geq \int_1^n \log x ~ dx
\\
&= \Big[ x\log x - x \Big]_1^n \\
&= n\log n - n + 1
\end{align}
from which you can deduce
$$\log n! - \frac{n}{2}
\log n \geqslant \frac{n}{2} \Big( \log n - 2 \Big) + 1.$$
The right hand side is clearly positive for all sufficiently large $n$ because $\log n$ is unbonded.  A little calculation can show the threshold is $n \geqslant 5$.  Taking exponentials it follows,
$$n! \geqslant n^{n/2}\quad\text{when}\quad n \geqslant 5.$$
The inequality thus derived gives a sufficient constraint on $n$.  It also shows the inequality is strict for large enough $n$.  A little more calculation can show it also holds when $n=1,2,3$ and $4$.  Our choice of step function was too coarse to identify the inequality remains valid in these cases.
A: Your approach is basically OK, but it really should include a proof of the key inequality $n!\gt n^{n/2}$ (for $n\gt2$). Here's a proof by induction, starting with $2!=2^{2/2}$ and using the inequality $(1+1/n)^n\lt e\lt3$.
If $n!\ge n^{n/2}$, then the strict inequality $(n+1)!\gt(n+1)^{(n+1)/2}$ follows:
$$\begin{align}
(n+1)!
&=(n+1)n!\ge(n+1)n^{n/2}\\
&=(n+1)^{(n+1)/2}\left(n+1\over\left(1+{1\over    n}\right)^n\right)^{1/2}\\
&\gt(n+1)^{(n+1)/2}\left(2+1\over3 \right)^{1/2}\\
&=(n+1)^{(n+1)/2}
\end{align}$$
Note, the assumption $n\ge2$ comes into play in the numerator replacement $n+1\ge2+1$; everything else holds even for $n=1$; in particular the strict inequality $(1+1/n)^n\lt3$ holds for $n\ge1$. Of course, one can (and maybe should) now insist on a proof of $(1+1/n)^n\lt 3$, but I'll take that inequality as known.
A: The problem I see in your proof is the first inequality $n! >n^{n/2}$, I don't think it's correct even for $n$ sufficiently large because $\frac{n!}{n^n} \rightarrow 0$ (although this is a bit different from what you say).
If I had to give a simple and basic justification for that I'd do de following.
Consider any fixed $m \in \mathbb{N}$, then for $n > m , n \in \mathbb{N}$ we have
$$n!=m! \cdot (m+1) \cdots (n-1)n > m! \cdot m \cdot m \cdots m = m! \cdot m^{n-m}$$
Takin n-roots you get $\sqrt[n]{n!} > \sqrt{m!} \cdot m^{1-\frac{m}{n}} > m^{1-\frac{m}{n}}$. Now for $n$ big enough you have $\frac{m}{n} <\frac{1}{2}$ and thus $1-\frac{m}{n} > \frac{1}{2}$. So for $n$ big enough you deduce $\sqrt[n]{n!} > \sqrt{m}$
And since you can make $\sqrt{m}$ as large as you want, you deduce $\sqrt[n]{n!}\rightarrow +\infty$
Keep in mind that these approximations are really really suboptimal, but still, they get the job done.
A: Assume $(n!)^\frac{1}{n}$ is bounded above by some $K$, i.e.
$$(n!)^\frac{1}{n}\le K$$
Then $$n!\le K^n$$
Let $n=K^2+K$. Then $$n!=1\ \cdot 2 \cdots K\cdot (K+1) \cdots K^2 \cdot (K^2+1) \cdots (K^2+K)$$
The product of the first $K$ terms with the last $K$ terms is greater than $K^{2K}$, and every remaining term is greater than $K$.
We have $K^2+K$ terms in total, each one (or a pairing of) is greater than $K$ (or $K^2$), hence $(K^2+K)!\gt K^{K^2+K}$, a contradiction.
