Simple(r) proof that $\pi(2^n)\geq n$? We can clearly prove that $\pi(2^n)\geq n$ with Bertrand's postulate, but that seems like overkill. 
Is there any simpler way one can prove that $\pi(2^n)\geq n$?
Note: $\pi(m)$ is the prime counting function - the number of primes $\leq m$.
 A: I am going to answer my own question in the negative, and say that there is no apparent commonly known "simple" proof. 
The existence and public knowledge of a nice proof for $\pi(2^n)\geq \frac{n}{2}$ would indicate if there was a similar-length similar-level proof for $\pi(2^n)\geq n$, someone would know of it.

From "Proofs From The Book," there is an easy proof that:
$$\pi(x)+1\geq \log_e x$$
This proof is easier than the proof that $\pi(2^n)\geq \frac{n}{2}.$ 
Specifically, if $n\leq x<n+1$ then:
$$\log_e(x)\leq \sum_{m=1}^{n}\frac{1}{m} \leq \prod_{i=1}^{\pi(x)}\frac{1}{1-1/p_i}=\prod_{i=1}^{\pi(x)}\frac{p_i}{p_i-1}\leq \prod_{i=1}^{\pi(x)}\frac{i+1}{i}=\pi(x)+1$$
This lets you prove, for example, that $\pi(x)\geq \log_3(x)$ for all $x$ simply by checking finitely many values - specifically, the $x$ such that $\log_e x -1 \leq \log_3 x,$ which amounts to $x\leq 68922.$
This doesn't get us much close to the desired $\pi(x)\geq\log_2(x).$
This proof does give us $\log_4(x)\leq \pi(x)$ just by checking the cases $x\leq 55$ since $\log_e(x)-1\geq \log_4(x)$ for $x>55.$
