TITLE: What is the simplest lower bound on prime counting functions proof wise?
QUESTION [18 upvotes]: I am trying to understand how a lower bound can exist on the prime counting function, and to begin this process of educating myself I am trying to find a simple complete proof, that does not hinge on another more complex theory?

REPLY [2 votes]: Here is an answer that uses the Fundamental Theorem of Arithmetic. Let $\pi(n)$ be the number of primes less than or equal to $n$. For each natural $k$ less than $n$, the exponent of the primes in the prime factorization of $k$ can be at most $\log_2(n)$. 
Since there are $\pi(n)$ total primes that can be used in the prime factorization of $k$, we have 
$$(\log_2(n) + 1)^{\pi(n)} \ge n \implies \pi(n) \ge \frac{\log(n)}{\log(\log_2(n) + 1)} \approx \frac{\log(n)}{\log \log(n)}.$$
Interestingly, exponentiating this bound gives you the PNT.