I'm trying to show that $\omega(n)$ is less than $\frac{\log n}{\log(\log n)}$ as it's stated without proof in an analytic number theory text. It's a corollary of the PNT, but I want to not use that as it's too strong a result to be enlightening.
My thoughts so far: look only at the primorials. It's easy to show that it's less than $\log_2n$ by a contradiction argument and a weak lower bound on $k\#$ of $2^k$
$\log(\log n)$ immediately makes me think of rooting something of size $\log n$ so what if tried to strengthen the lower bound on $k\#$ since if we were to say divide $\omega(n) = n < \log n$ by two, we could write it as $p_{1}...p_{1} p_{k/2}... p_{k/2}$ Then by looking at each "prime power" we want to make an argument that a certain number of those primes have to be $1$. But the contradiction argument doesn't work in this case.
Hints are greatly appreciated, no pure answers please.
I believe I have a solution to this problem: Let $n \in \mathbb{N}$. Then since n has a prime factorization, we have $ n \ge k\# = p_1, ... ,p_k \ge k! \ge (\frac{k}{2})^{\frac{k}{2}}$. Clearly $\omega(n)=k$. Then since $\log$ is a monotonically increasing function, we have $\log n \ge \frac{k}{2} \log \frac{k}{2} \ge \frac{k}{2} \log k$ and since $\log \neq 0$ we have $\frac{\log n}{\log \omega(n)} \ge \omega(n)$. Since $\omega(n) \le \log n$ by the monotonicity of $\log$, we obtain $\omega(n) \le \frac{\log n}{\log \log n}$.