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 $n > 1$ then:
$\log(n)  <  n$
$\log(\log(n)) <  \log(n)$
But I can't figure out what my C should be if this is the correct answer?

If $0<f(n)<g(n)$ then $c=1$ works to show $f(n)=O(g(n))$.
