A trivial question about the $n^{\text{th}}$ prime number I would like to ask for a little help about a very trivial question from the number theory.

If $p_{n}$ is the $n^{\text{th}}$ prime number in the ascending sequence of prime numbers, show that $$p_{n}\sim n \log(n).$$

I am pretty sure, we should use the prime number theorem, which says that $$\mathbb{P}(x) = \sum_{p\leq x}1 \sim \frac{x}{\log(x)} $$ where $p$ is a prime number.
I would be glad if someone could help me with the idea. Thanks in advance!
 A: As Tomas noted: assuming
$$ \sum_{p \leq x} 1 \sim \frac{x}{\log (x)} $$
and using that
$$ \sum_{p \leq p_n} 1 = n $$
we can solve 
$$ \frac{p_n}{\log (p_n)} \sim n \implies p_n \sim n \log n $$

Edit Since there are some concerns about this last step, here I give a slightly clumsy derivation. 
First we have that $p_n / \log p_n  = n + o(n)$ from the prime number theorem. 
Second we have that 
$$ \frac{n \log n}{\log (n\log n)} = n ( 1 - \frac{\log\log n}{\log n + \log\log n}) = n + o(n) $$
Combining the two statements we have
$$ \frac{p_n}{\log p_n} = \frac{n\log n}{\log (n\log n)} + o(n) \tag{*}$$
Now consider the function $g:x\mapsto x / \log x$. Its derivative is 
$$ g'(x) = \frac1{\log x} - \frac{1}{(\log x)^2} $$
which is positive if $x > e$. So over the domain $(e,\infty)$ the function is invertible, with range $(e,\infty)$.  
Now, consider $g^{-1}( g(n\log n) + o(n))$. Since $g(n \log n / 2) < g(n\log n) + o(n) < g(2n \log n)$ for sufficiently large $n$, we have that in the relevant interval 
$$\frac{1}{2\log n} < g' < \frac{1}{\log n}$$
so that 
$$ g^{-1}(g(n\log n) + o(n)) = n\log n + o(n\log n) $$
So applying $g^{-1}$ to both sides of (*) we get that
$$ p_n = n\log n + o(n\log n)$$
or
$$ p_n \sim n\log n$$
A: At the lowest level, this isn't a question about prime numbers at all; instead, it's a question about finding the inverse to the function $f(x) = \frac{x}{\log x}$.  A classic method for (asymptotically) approximating inverses is bootstrapping: start with some reasonable guess, 'plug it in' to the functional equation (in some form) to get a better approximation, and continue until your approximation stops getting better (or you get enough terms).
In this case, we want to find a $g(n)$ such that $n = \frac{g(n)}{\log g(n)}$; rearranging this gives $g(n) = n\log g(n)$, which seems like a fine place to start the bootstrapping.  A reasonable first guess is simply $g_0(n)=n$ (note that starting with a constant $g_0()$ would lead to a linear guess on the first iteration); plugging this in to the equation yields, successively:
$$g_1(n) = n\log g_0(n) = n\log n$$
$$g_2(n) = n\log g_1(n) = n\log(n\log n) = n(\log n+\log\log n)$$
$$g_3(n) = n\log\left(n\log n+n\log\log n\right) = n\log\left(n\log n\cdot\left(1+\frac{\log\log n}{\log n}\right)\right)\\=n\left(\log n+\log\log n+\log\left(1+\frac{\log\log n}{\log n}\right)\right)\\=n\left(\log n+\log\log n+\frac{\log\log n}{\log n}+o\left(\frac{\log\log n}{\log n}\right)\right)$$
In the case to hand, of course, there are error terms in the original approximation $\pi(n) = \frac{n}{\log n}$ that make the higher-order terms here less relevant, but a slightly more careful bootstrapping will show that the top-order $\theta(n\log n)$ term survives unscathed; the exact expression, as per Wikipedia, is:
$$\frac{p_n}{n} = \log n+\log\log n-1+O\left(\frac{\log\log n}{\log n}\right)$$
