It is said that the Euler product $$\prod_p \frac{1}{1-p^{-s}}$$ diverges as $s \to 1^+$ proves we can't find constants $C$,$\theta$ with $\theta < 1$ such that $\pi(x) < C x^\theta$ because that would imply that the product converges for $s > \theta$.
I don't understand this deduction at all, how is the fact about the prime counting function concluded?
Taking logs of the Euler product, expanding out the resulting series, and comparing all the higher-order terms (i.e. terms involving $p^{-ns}$ for $n > 2$) to the convergent series $\sum_{m = 1}^{\infty} m^{-2s}$, we find that the divergence as $s \to 1^{+}$ is equivalent to the divergence of the series $\sum_p p^{-s}$ as $s \to 1^{+}$. Now if $\pi(x) < C x^{\theta}$, then the $n$th prime would have size $\sim n^{1/\theta}$, and so this series would behave like $\sum_{n = 1}^{\infty} n^{- s/\theta},$ which converges as $s \to 1^+$ (because $1/\theta > 1$), a contradiction.