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.
The idea is that you use induction, and that you expand on Euclid's argument that there are infinitely many primes to show that $p_n$ is bounded above by $2^{2^n}$.
Suppose $p_i \leq 2^{2^i}$ for all $i \leq n$. Now $p_1 \cdot p_2 \cdot \ldots \cdot p_n + 1$ is not divisible by any of the primes $p_1, \ldots, p_n$, so
$$p_{n+1} \leq \prod_{i=1}^n \ p_i + 1$$
Substituting the earlier bounds on $p_i$ and using $\sum_{i=1}^n 2^i = 2^{n+1} - 2$, we get
$$p_{n+1} \leq \left(\prod_{i=1}^n \ 2^{2^i}\right) + 1 = \left(2^{\sum_{i=1}^n 2^i}\right) + 1 = 2^{2^{n+1} - 2} + 1 \leq 2^{2^{n+1}}$$
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.
Here we look at a smaller, and as I feel, more convenient bound:
Assume each prime would be the double of the previous, we had, beginning with the first prime q=2 the sequence $ \small q, 2q, 4q, 8q, \ldots 2^n q , \ldots...$ and the n'th prime were bounded by $ \small q2^n $ which is much smaller than the given $ \small 2^{2^n} $ . Now if we recall that between n and 2n there is always a prime, thus the distance to the next is even smaller than n , then this is sufficient for the proof in general and we need only look at the initial cases for possible exceptions.
