Finding asymptotic relationship between: $\frac {\log n}{\log\log n} \overset{?} = (\log (n-\log n))$ 
Given $f(n)=\frac {\log n}{\log\log n} , g(n)= (\log (n-\log n))$, what is the relationship between them $f(n)=K (g(n))$ where "K" could be $\Omega,\Theta,O$

I thought of taking a log to both sides and see what we get:
$\log\frac {\log n}{\log(\log n)}= \log(\log n) -\log[\log(\log n)]  \overset{?} = c\log(\log (n-\log n))$
It looks like the RHS is smaller than: $\log(\log (n-\log n)) \le \log(\log n)$
And since $\log[\log(\log n)] < \log(\log n)$ then $\log(\log (n-\log n)) \le \log(\log n) - \log[\log(\log  n)] $
But it's actually $O$, and I can't find a way to show it...

Hint:
$$\lim\limits_{n\to \infty}\frac{f(n)}{g(n)}=
\lim\limits_{n\to \infty}\frac{\log n\bigg/\log(\log n)}{\log(n-\log n)}=\lim\limits_{n\to \infty}\frac{\log n}{\log(\log n)\log(n-\log n)}=
\lim\limits_{n\to \infty}\frac{\log n}{\bigg(\log n+\log(1-\frac{\log n}{n})\bigg)\log(\log n)}
=\lim\limits_{n\to \infty}\frac{\log n}{\log(\log n)\log n}=
\lim\limits_{n\to \infty}\frac{1}{\log(\log n)}=0$$
