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 $\ln({p_n})$?

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
