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Didn't Euler prove that $\sum 1/p$ diverges? This proves that $\pi(N)$ is infinitely often larger than $N^{1-\epsilon}$: If there were only $N^{1-\epsilon}$ primes less than $N$, then there are at most $2^{k (1-\epsilon)}$ primes between $2^{k-1}$ and $2^k$. So we can bound $\sum 1/p$ above by $\sum 2^{k(1-\epsilon)}/2^{k-1} = 2 \sum 2^{-k \epsilon}$, which converges.

UPDATE: I should clarify that I see no way to get from here to the stronger statement that $\pi(N)$ is greater than $N^{1 - \epsilon}$ for all sufficiently large $N$.

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Didn't Euler prove that $\sum 1/p$ diverges? This proves that $\pi(N)$ is infinitely often larger than $N^{1-\epsilon}$: If there were only $N^{1-\epsilon}$ primes less than $N$, then there are at most $2^{k (1-\epsilon)}$ primes between $2^{k-1}$ and $2^k$. So we can bound $\sum 1/p$ above by $\sum 2^{k(1-\epsilon)}/2^{k-1}= 2^{k(1-\epsilon)}/2^{k-1} = 2 \sum 2^{-k \epsilon}$, which converges.

Of course, rigorous notions of convergence didn't exist in Euler's day, but I think his proof could be turned into a modern one.

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Didn't Euler prove that $\sum 1/p$ diverges? If there were only $N^{1-\epsilon}$ primes less than $N$, then there are at most $2^{k (1-\epsilon)}$ primes between $2^{k-1}$ and $2^k$. So we can bound $\sum 1/p$ above by $\sum 2^{k(1-\epsilon)}/2^{k-1}= 2 \sum 2^{-k \epsilon}$, which converges.

Of course, rigorous notions of convergence didn't exist in Euler's day, but I think his proof could be turned into a modern one.