What are the current best asymptotic bounds on $\pi^{-1}(x)$, where $\pi(x)$ denotes the prime counting function (number of primes at most $x$)?
In other words, I am curious about the state of the art for estimating the $n^{th}$ prime. From the prime number theorem, it seems clear that $\pi^{-1}(x)=\Theta(x \log x)$. Can someone point me in the direction of literature that answers the question? Please excuse my inability to find such literature myself...
Here is how we invert the prime counting function $\pi(x)$ to estimate the n-th prime $p_n$. Let $g(u),u \ge 2$ be positive, continuous increasing function and let $f(x)$ be defined by
$$ x = \int_{2}^{f(x)} \frac{du}{g(u)} $$
Then
$$ f(x) = \int_{2}^{x} g(f(u))du + f(2) $$
Take $g(u) = \ln u$ so our first estimate of $f(x) is $that $f(x) \sim x\ln x$. Now use this estimate in the above integral to obtain the second estimate for $f(x)$. You can repeat this iterative process $k$ times to obtain the asymptotic expansion of $f(x)$ with an error $O(x/(\ln x)^{-k})$.
Now from the Prime number theorem, we have
$$ \pi(x) \sim \int_{2}^{x} \frac{du}{\ln(u)} $$
Taking $x = p_n$, the n-th prime, the above method says that the asymptotic expansion of $p_n$ is $f(n)$. Thus we obtain
$$ p_n \sim n\ln n + n\ln\ln n - n + \ldots $$
Assuming the Riemann hypothesis $$|p_n -\text{li}^{-1}(n) |\le \pi^{-1} \sqrt{n} (\log n)^{5/2}, \qquad n \ge 11$$ This follows from known bounds for $\pi(x)$ due to Schoenfeld.
You may see the details in the paper https://arxiv.org/abs/1203.5413
Since "the state of the art" slightly changed since 2013, I am adding this new result on lower and upper bounds on $p_n$. (This is complementary to the answers given earlier.)
Christian Axler proved in https://arxiv.org/abs/1706.03651 that:
(1) For every $n\ge46254381$, $${\small p_n < n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 10.667 \over 2\log^2 n} \Big). } $$
(2) For every $n\ge2$, $${\small p_n > n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 11.508 \over 2\log^2 n} \Big). } $$
[I think this is morally the same as Nilotpal Sinha's answer, but it only uses asymptotic estimates and no integrals, so I thought I'd post it here in case someone found it useful or interesting.]
Start with the version of the Prime Number Theorem which says that $\pi(x) \sim x / \ln(x)$. Let $p_n$ denote the $n$-th prime, so in particular $\pi(p_n) = n$, and substitute in $p_n$ in the asymptotic estimate for $\pi(x)$ to get $n = \pi(p_n) \sim p_n / \ln(p_n)$, or equivalently
$p_n \sim n \ln p_n$
(If you're being nitpicky this works because the sequence $p_n$ is monotonic and unbounded.) Now take logs of both sides (again, if you're being nitpicky, this works because both $n$ and $p_n / \ln(p_n)$ are unbounded) and multiply by $n$ to get:
$n \ln p_n \sim n \ln(n) + n\ln\ln(p_n)$.
At this point, if you want just the leading term, you can show easily that $\lim_{n \to \infty} n \ln\ln(p_n) / n\ln(n) = 0$, so that in fact $p_n \sim n \ln p_n \sim n \ln n$, or you can probably keep substituting in to get finer and finer error terms.
