Suppose that the \(n\)th prime number is given by \(p_n\) and the number of primes that do not exceed \(n\) is \(\pi(n)\). Let two functions \(f\) and \(g\) be asymptotic to each other, denoted by \(f\sim g\), iff \(\lim_{x\rightarrow\infty}{\frac{f(x)}{g(x)}}=1\). The prime number theorem states \[\pi(x)\sim \frac{x}{\log{x}}.\] This statement happens to be equivalent to \(p_n\sim x\log{x}\). I will show this equivalence, using the argument made in An Introduction to the Theory of Numbers, though with much of the gaps and leaps in reasoning filled in by myself.

The first thing to note is that \(p_n\) and \(\pi(n)\) are inverse functions. That is, \(\pi(p_n)=n\) and \(p_{\pi(n)}=n\). Now, we consider the function \[y=\frac{x}{\log{x}}.\] Taking the logarithm of both sides, we obtain, \[\log{y}=\log{x}-\log{\log{x}}.\] Now, we show that \(\log{\log{x}}=o(\log{x})\). By L’Hôpital’s rule, \[\lim_{x\rightarrow\infty}{\frac{\log{\log{x}}}{\log{x}}}=\lim_{x\rightarrow\infty}{\frac{1}{\log{x}}}=0,\] as desired. So now, we can divide the logarithm of \(y\) by the logarithm of \(x\) to obtain \[\frac{\log{y}}{\log{x}}=1-\frac{\log{\log{x}}}{\log{x}}.\] In the limit, this approaches 1, so that \(\log{y}\sim \log{x}\). Hence, when we have \[x=y\log{x},\] we have that the LHS is the inverse of \(y\), but \(\log{x}\) is asymptotic to \(\log{y}\), so the inverse of \(y\) is asymptotic to \(y\log{y}\).

Now, it suffices to show that for two divergent functions \(f\sim g\), we must also have \(f^{-1}\sim g^{-1}\). Observe that \[f^{-1}(f(x))=x\Rightarrow\lim_{x\rightarrow\infty}{\frac{f^{-1}(f(x))}{x}}=1.\] Likewise, we have \[\lim_{x\rightarrow\infty}{\frac{g^{-1}(g(x))}{x}}=1.\] But since \(f\sim g\), we can replace \(g\) with \(f\) in the limit to obtain \[\lim_{x\rightarrow\infty}{\frac{g^{-1}(f(x))}{x}}=1.\] Dividing our two limits, we obtain \[\lim_{x\rightarrow\infty}{\frac{f^{-1}(f(x))}{g^{-1}(f(x))}}=1,\] and since \(f\) is divergent, we can write this as \[\lim_{f(x)\rightarrow\infty}{\frac{f^{-1}(f(x))}{g^{-1}(f(x))}}=1,\] so \(f^{-1}\sim g^{-1}\), as desired.

Hence, since \(\pi(n)\sim y\), the inverse of \(y\) is asymptotic to \(n\log{n}\), and \(p_n\) is the inverse of \(\pi(n)\), we have \[\boxed{p_n\sim n\log{n}}.\]