TITLE: How to show $p_n$ $\leq$ $2^{2^n}$?
QUESTION [14 upvotes]: Let $p_n$ be the $n_{th}$ prime (e.g. $p_1 = 2$; $p_2 = 3$; $p_3 = 5$). Show that $p_n \leq 2^{2^n}$ for all $n$.

I don't see how I can approximate the value of $p_n$. Do I need something like prime number theorem? But the course didn't even teach that yet.

REPLY [18 votes]: It is enough to prove that for any positive integer $n$, there are at least $n$ primes that are $\le 2^{2^n}$. We will show that in fact $2^{2^n}-1$ always has at least $n$ distinct prime divisors. (All of these are necessarily $<2^{2^n}$.) 
Suppose we know that $2^{2^k}-1$ has at least $k$ distinct prime divisors.  Then we must show that $2^{2^{k+1}}-1$ has at least $k+1$ distinct prime divisors. 
Note that
$$2^{2^{k+1}}-1=(2^{2^k}-1)(2^{2^k}+1).$$
The numbers $2^{2^k}-1$ and $2^{2^k}+1$ are relatively prime.  This is because any number that divides both of them must divide their difference, which is $2$.  But each of 
$2^{2^k}-1$ and $2^{2^k}+1$ is odd, so the greatest integer that divides both of them cannot be $2$, and therefore must be $1$.
Thus the prime divisors of $2^{2^{k+1}}-1$ are the prime divisors of $2^{2^k}-1$ (of which there are at least $k$), together with the prime divisors of $2^{2^k}+1$ (of which there is at least $1$), for a total of at least $k+1$.
Comment: The above argument also tells us that for any positive integer $n$, there are at least $n$ primes.  This gives us a proof of the infinitude of the primes which is a little different from the usual "Euclid" proof.