Prove log(log(n)) is Big-O (log(n)) I need to show that:

$g(n) = \log(\log(n))= O(\log(n))$

This is what I have so far:

Choose $k = 1$

Suppose $n &gt; 1$ then:

$\log(n)  &lt;  n$

$\log(\log(n)) &lt;  \log(n)$

But I can't figure out what my C should be if this is the correct answer?
If $0&lt;f(n)&lt;g(n)$ then $c=1$ works to show $f(n)=O(g(n))$.
If $0&lt;f(n)&lt;g(n)$ then $c=1$ works to show $f(n)=O(g(n))$.