Problem
Prove that $\pi(x) > \log x - 1$.
Progress
Based on a hint and very elementary methods, I got that $$ \prod_{p \leq x} (1-p^{-1})^{-1} \leq \prod_{k=2}^{\pi(x)+1} (1-k^{-1})^{-1}. $$ The second product telescopes to $\pi(x)+1$, and based on numerical calculations, I should be able to show that $\log x$ is less than this product. In fact, using this trivial bound $$ \log x = \int_1^x {dt \over t} < \sum_{k=1}^{[x]+1} {1 \over k}, $$ seems to hold for $x \geq 5$. The remaining cases could be checked separately. So is there a simple way to show that $$ \sum_{k=1}^{[x]+1} {1 \over k} < \prod_{p \leq x} (1-p^{-1})^{-1} $$ for $x \geq 5$? It only needs to be verified for positive integers, so I thought induction might work, but I'm not seeing how to deal with a sum and a product together. I also thought about expanding $\log x$ into a series another way, but its power series doesn't converge where we would need it to. In fact, I'm fairly confident this is the right approach, because of the close comparison between $\log x$ and these partial sums of the harmonic series. One approach from here is to do something like this, but with a finite number of terms, it seems to get really messy really fast.
Notes: $\log$ is natural, $[x]$ is the floor function, and $p$ ranges over primes.