Let $\operatorname{LCM}[n]:=\operatorname{lcm}\{1,\ldots,n\}$. It is easy to verify $\operatorname{LCM}[n]\geq 2^{\pi(n)}$, where $\pi(n)$ counts the number of distinct primes up to $n$.
But how can I prove the bound $\operatorname{LCM}[n]\geq (\sqrt{n})^{\pi(n)}$?
For the bound $2^{\pi(n)}$, I used the identity:
$\operatorname{LCM}[n]=\!\!\!\prod_{p\in\mathbb{P},p\leq n} p ^{\lfloor \log_p(n) \rfloor}$
But this doesn't work with the lower bound of $\operatorname{LCM}[n]\geq (\sqrt{n})^{\pi(n)}$. Can someone give me a hint?