The n-th prime is less than $n^2$? [duplicate]
Let $p_n$ be the n-th prime number, e.g. $p_1=2,p_2=3,p_3=5$. How do I show that for all $n>1$, $p_n<n^2$?
In Zagier's the first 50 million prime numbers a very elementary proof is given that for $n > 200$ we have
$$\pi(n) \ge \frac23 \frac{n}{\log n}$$
where $\pi(x)$ is the number of primes below $x$ (as well as a bound in the other direction). In fact, it already holds for $n \ge 3$, as can be directly checked.
Suppose that $p_n > n^2$, then for this $n$ we have $\pi(n^2) < n$, but that violates the bound already for $n = 5$.