Htykuut

Quoting from the Wikipedia article:

As a consequence of the prime number theorem, one gets an asymptotic expression for the $ n $th prime number, denoted by $ p_n $:
$$ p_n sim n log n.$$

Can you explain how we get this approximate expression? I understand the number of prime numbers less than or equal to an integer $ N $, denoted as $ pi(N) $ is approximately:
$$
pi(N) sim frac{N}{log N}.
$$

If the $ n $ prime number is approximately $ n log n $, it implies that there are approximately $ n $ prime numbers less than or equal to $ n log n $, i.e.,
$$
pi(n log n) sim n.
$$
But substituting $ N $ with $ n log n $ in the formula of $pi(N)$ we get
$$
pi(N) sim frac{N}{log N} = frac{n log n}{log (n log n)} notsim n.
$$
What am I doing wrong? How can I show that the statement quoted from Wikipedia is correct?

$$log(nlog n)=log n+loglog n=(1+o(1))log n$$
therefore
$$frac{nlog n}{log(nlog n)}=frac{nlog n}{(1+o(1))log n}
=frac{n}{(1+o(1))}sim n.$$

It seems you are very new to prime number theory hence I am not sure if you understand the little $o$ notations. Hence I present @Lord Shark's answer in layman's terms.

$$
log (nlog n) = log n + loglog n
$$

But $loglog n$ is exponentially smaller compared to $log n$ i.e. it can be ignore compared to $log n$. Hence as an approximation, we have

$$
log (nlog n) sim log n
$$

With this approximation

$$
pi (nlog n) sim frac{nlog n}{log (nlog n)} sim frac{nlog n}{log n} sim n
$$

The smallest values of $x$ for which $pi(x) = n$ is $x = p_n$ hence we have

$$
pi (p_n) = n sim pi(nlog n)
$$

or $p_n sim nlog n$.