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 $\ln({p_n})$?
 A: Made a jpeg, see if it comes out readable: Theorem 3 says that putting in the extra $\log \log n$ term is quite good.

Theorem 1. We have $$\frac{x}{\log x}\left(1 + \frac{1}{2 \log x}\right) < \pi(x)\qquad \text{for}\,59\leqq x, \tag{3.1}\label{3.1}$$ $$\pi(x) < \frac{x}{\log x}\left(1 + \frac{3}{2 \log x}\right) \qquad \text{for}\,1 < x. \tag{3.2}\label{3.2}$$
Theorem 2. We have $$x/(\log x - \tfrac{1}{2}) < \pi(x) \qquad\text{for}\,67 \leqq x, \tag{3.3}\label{3.3}$$ $$\pi(x) < x/(\log x - \tfrac{3}{2})\qquad \text{for}\,e^{3/2} < x$$ (and hence for $4.48169 \leqq x$).
Corollary 1. We have $$x/\log x < \pi(x) \qquad\text{for}\,17\leqq x, \tag{3.5}\label{3.5}$$ $$\pi(x) < 1.25506 x/\log x \qquad\text{for}\,1 < x. \tag{3.6}\label{3.6}$$
Corollary 2. For $1 < x < 113$ and for $113.6 \leqq x$ $$\pi(x) < 5x/(4 \log x). \tag{3.7}\label{3.7}$$
Corollary 3. We have $$3x/(5 \log x) < \pi(2x) - \pi(x) \qquad\text{for}\,20\tfrac{1}{2} \leqq x, \tag{3.8}\label{3.8}$$ $$0 < \pi(2x) - \pi(x) < 7x/(5 \log x) \qquad\text{for}\, 1 < x. \tag{3.9}\label{3.9}$$
For the ranges of $x$ for which these corollaries do not follow directly from the theorem, they can be verified by reference to Lehmer's table of primes [10]. A similar remark applies to all corollaries of this section unless a proof is indicated.
The inequality \eqref{3.8} improves a result of Finsler [3]. The left side of \eqref{3.9} is just the classic result, conjectured by Bertrand (and known as Bertrand's Postulate) and proved in Tchebichef [14], that therre is at least one prime between $x$ and $2x$. The right side of \eqref{3.9} gives a result of Finsler [3], with Finsler's integral $n$ replaced by our real $x$. Finsler's elementary proofs are reproduced in Trost [15] on p. 58. The relation \eqref{3.12} below states a result of Rosser [11].
Theorem 3. We have $$n(\log n + \log\log n - \tfrac{3}{2}) < p_n \qquad\text{for}\,2\leqq n, \tag{3.10}\label{3.10}$$ $$p_n < n(\log n + \log\log n - \tfrac{1}{2})\qquad\text{for}\,20\leqq n. \tag{3.11}\label{3.11}$$
Corollary. We have $$n \log n < p_n \qquad\text{for}\, 1 \leqq n, \tag{3.12}\label{3.12}$$ $$p_n < n(\log n + \log\log n)\qquad\text{for}\,6\leqq n. \tag{3.13}\label{3.13}$$

That looks pretty good. From https://projecteuclid.org/euclid.ijm/1255631807
A: An elementary proof. Let $p_n=k_n n \ln n.$ We have $$1\sim \frac {n-1}{n}=\frac {\pi (p_n)}{n}\sim \frac {p_n}{n\ln p_n}=$$ $$=\frac {k_n n\ln n}{n\ln k_n+n\ln n n+n\ln\ln n}=$$ $$=\frac {k_n}{\frac {\ln k_n}{\ln n}+1+\frac {\ln\ln n}{\ln n}}=^{def}W_n.$$
(1).  When  $k_n\ge 1$ we have  $W_n= \frac {k_n}{o(k_n)+1+o(1)}.$  Since $W_n\sim 1$, this implies $ \lim\sup_{n\to\infty}k_n\le 1.$
(2a). If $|\ln k_n|=o(\ln n)$ then when $k_n<1$ we have $W_n=\frac {k_n}{1+o(1)},$ which, since $W_n\sim 1$ and $\lim\sup_{
n\to\infty}k_n\le 1$ from (1), implies $\lim\inf_{n\to\infty}k_n\ge 1.$
(2b).The other alternative $|\ln k_n|\ne o(\ln n)$ for $k_n<1$ would imply (since $\lim\sup_{n\to\infty}k_n\le 1$) that for some $c>0$  there are infinitely many $n$ for which $(-\ln k_n)=|\ln k_n|>c\ln n$ and hence $k_n<n^{-c}.$ But for such $n$ we would have $p_n=k_n n \ln n<n^{1-c}\ln n,$ and we have $\ln n<n^c$ for all sufficiently large $n,$ so we would have $p_n<n$ for some $n\ge 1$, which is absurd. So this alternative is impossible.
Remark. $W_n\sim 1$ is by itself insufficient to prove $k_n\to 1.$ We must, as in (2b), use the relation between $k_n$ and $p_n$ and the fact that $p_n>n. $
A: 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.
