Why is $p_n \sim n\ln(n)$? I know that the prime number theorem states that the number of primes less than $x$ is approximately $\frac{x}{\ln(x)}$. However, why does this mean that $p_n \sim n\ln(n)$? (where $p_n$ is the $n$-th prime). If we replace $x$ with $p_n$ in the original equation, we have that $\pi(p_n) \ln({p_n})\sim p_n$, and $\pi(p_n)$ is just $n$, but what about the $\ln({p_n})$?

If viewed as an expectation
$\pi(n)=\frac{n}{\ln(n)}$
reads 'throw an event which has probability $\frac{1}{\ln n}$ $n$ times'.
So a number is prime with probability $\frac{1}{\ln n}$. So the expected number of throws to get a prime is $\ln n$.
So the $n^{th}$ prime expects $n\ln n$ throws.
