Prime counting function inequality It seems to be the case that for any integer $x > 1 $, we have $ x \leq 2^{\pi(x)} $
I'm not sure whether this is obviously true, annoyingly false or difficult to prove. I'm hoping it's the former, and that I'm just being dim. I'm thinking along the lines of prime factorisation of $x$, but I can't see it. Any help would be greatly appreciated.
Thanks
 A: Suppose that $x$ is minimal such that $x>2^{\pi(x)}$. Let $y=\lceil x/2 \rceil$; then $2y\ge x>2y-2$. Bertrand’s postulate in its stronger form ensures that there is a prime between $y$ and $2y-2$ and hence between $y$ and $x$, so $\pi(x)\ge\pi(y)+1$, and therefore $2^{\pi(x)}\ge 2\cdot 2^{\pi(y)}\ge 2y = x$, which is a contradiction.
A: In "A Classical Introduction to Modern Number Theory" by Ireland and Rosen, a very close bound is obtained by completely elementary means.
Assum $p_n \le x < p_{n+1}$, where $p_i$ denotes the $i$'th prime. By definition, $\pi(x) = n$.
Considers the numbers up to $x$: $\{1,2,\cdots, x \}$. They are divisible only by primes among $\{ p_1, p_2, \cdots, p_{n} \}$. If you decompose those numbers into a square part and square-free part, i.e. write $a = rs^2$ where $r$ is a product of distinct primes and $s$ is a square, we see that there are 2 constraints on $r,s$:


*

*Since $r$ is square-free, it must be a product of distinct primes among the first $n$ primes. There are only $2^{n}$ options for $r$.

*Since $s^2 \le a$, we must have $s \le \sqrt{a} \le \sqrt{x}$, so there are at most $\sqrt{x}$ options for $s$.
All in all, there are at most $2^n \times \sqrt{x}$ options for those numbers between $1$ and $x$:
$$ x \le 2^{n} \sqrt{x} = 2^{\pi(x)} \sqrt{x}$$
This shows that $\pi(x) \ge \log_{2} \sqrt{x} = \frac{\log_{2} x}{2}$. $\blacksquare$
Similarly, if we decompose those numbers into an $n$'th-power and an $n$-powerfree number, we find:
$$\pi(x) \ge \frac{\ln x (1-\frac{1}{n}) }{\ln n} $$
But $n=2$ already gives the best bound. I will think later about improving this to $\pi(x) \ge \log_2 x$ - I believe it is possible with a similar method.
