n >= (k-2)! implies k = O(log n / log log n)? Let n and k be some integers.  k > 3.
Let $n \geq (k-2)!$.
I have to show, that $k = \mathcal O\left(\frac{\log{n}}{\log{\log{n}}}\right)$
This statement was used in the article On vertex rankings of graphs and its relatives by I.Karpas et. al. I have to verify it for my bachelor project. The statement was marked as "easy to see" in the article. 
By using Stirling's approximation one can show that
\begin{equation*}
\begin{split}
\log{n} & \geq \log{\left(k-2\right)!}\\
& \geq \log{\sqrt{2\pi}} + \left(k-\frac{3}{2}\right)\log{\left(k-2\right)} - \left(k-2\right)\log{e}\\
& \geq \left(k-2\right)\log{\left(k-2\right)} - \left(k-2\right)\log{e}\\
& = \left(k-2\right)\left(\log{\left(k-2\right)} - \log{e}\right)\
\end{split}
\end{equation*}
That's actually all I could find out till now.
I was also trying to assume that $k \neq \mathcal O\left(\frac{\log{n}}{\log{\log{n}}}\right)$ and to show that $k > c\cdot\frac{\log{n}}{\log{\log{n}}}$ for all $c>0$ would lead to contradiction. 
Other possibility would be to show that, there exists $c>0$ such that
$$
\frac{\log{n}}{\log{\log{n}}} \geq \frac{\log{(k-2)!}}{\log{\log{(k-2)!}}} \geq\ldots\geq c\cdot k
$$
Unfortunately I fail on both approaches.
My question: Do you have any idea how to properly estimate the statements above or any other approaches to conclude the main statement? 
 A: Let $m=k-2$
\begin{equation*}
\begin{split}
\frac{\log{n}}{\log{\log{n}}} & \geq \frac{\log{m!}}{\log{\log{m!}}} \\
& \approx \frac{\log{\left(m^{m}\cdot e^{-m}\sqrt{2\pi\cdot m}\right)}}{\log{\left(\log{\left(m^{m}\cdot e^{-m}\sqrt{2\pi\cdot m}\right)}\right)}} \\
& = \frac{m\log{m} - m\log{e} + \frac{1}{2}\log{2\pi} + \frac{1}{2}\log{m}}{\log{\left(m\log{m} - m\log{e} + \frac{1}{2}\log{2\pi} + \frac{1}{2}\log{m}\right)}} \\
& = \frac{\left(m+\frac{1}{2}\right)\log{m} \overbrace{- m + \frac{1}{2}\log{2\pi}}^{\leq 0\text{ for } m > \frac{1}{2}\log{2\pi}}}{\log{\left(m\log{m} \underbrace{- m + \frac{1}{2}\log{2\pi} + \frac{1}{2}\log{m}}_{\leq m\log{m}}\right)}} \\
& \geq \frac{\left(m+\frac{1}{2}\right)\log{m}}{\log{\left(2m\log{m}\right)}} \\
& = \frac{\left(m+\frac{1}{2}\right)\log{m}}{\log{m} + \log{\left(2\log{m}\right)}} \\
& = \frac{\left(m+\frac{1}{2}\right)}{1 + \frac{\log{\left(2\log{m}\right)}}{\log{m}}} \\
& \geq \frac{m}{2} \\
& = \frac{k-2}{2}
\end{split}
\end{equation*}
A: I guess your answer is correct. Still, I think we can make it much shorter without using Stirling's approximation (please point out if there are errors).
Let $m=k-2$
We have $\log m!=\Theta(m\log m)$.
Hence there are positive constants $a,b$ such that $a~x\log x\leq \log x!\leq b~x\log x$
$$\frac{\log n}{\log\log n}\geq \frac{\log m!}{\log \log m!}\geq \frac{am\log m}{\log(bm\log m)}=a\frac{m\log m}{\log b+\log m+\log \log m}$$
Therefore,
$$m\leq \frac{1}{a}\frac{\log b+\log m+\log \log m}{\log m}\frac{\log n}{\log \log n}\leq c \frac{\log n}{\log \log n}$$ where $c$ is a positive constant (eg:$c=3/a$) assuming $m\geq b$. Hence $m=O(\log n/\log \log n)$
