Estimate of $n$th prime There is a result that if $p_n$ is the $n$th prime, then $p_n\sim n\log n$ as $n\rightarrow\infty$.
I wonder: Is it a direct consequence of the prime number theorem $\pi(x)\sim x/\log x$? The theorem says that there are approximately $n/\log n$ primes less than or equal to $n$. So there are approximately $n$ primes less than or equal to $n\log n$. So $p_n$ is approximately $n\log n$.
But I'm having trouble turning this into a formal argument. We have $\lim_{n\rightarrow\infty}\dfrac{\pi(n)\log n}{n}=1$. How would it show that $\lim_{n\rightarrow\infty}\dfrac{p_n}{n\log n}=1$?

Note that $\pi(p_n)=n$ for all positive integers $n$ and $p_{\pi(q)}=q$ for all primes $q$.  Thus
$$\lim_{n\to\infty}\left({n\log n\over p_n}\right)=\lim_{n\to\infty}\left({\pi(p_n)\log p_n\over p_n}\cdot{\log n\over\log p_n}\right)$$
Now
$$\lim_{n\to\infty}\left({\pi(p_n)\log p_n\over p_n}\right)=\lim_{p\to\infty}\left({\pi(p)\log p\over p}\right)=1$$
is just the Prime Number Theorem.  So it suffices to show that
$$\lim_{n\to\infty}\left({\log n\over\log p_n}\right)=1$$
Let's replace the sequence of integers $n$ with the sequence $\pi(q)$ where $q$ runs over the increasing sequence of primes.  Using the relation $p_{\pi(q)}=q$, this gives
$$\lim_{n\to\infty}\left({\log n\over\log p_n}\right)=\lim_{q\to\infty}\left({\log\pi(q)\over\log q}\right)=\lim_{q\to\infty}\left({\log\left({\pi(q)\log q\over q}\right)+\log q-\log\log q\over\log q}\right)=1$$
where the final limit evaluation again makes use of the Prime Number Theorem, this time for $\log\left({\pi(q)\log q\over q}\right)\approx0$.
