I'm trying to understand some particulars of the following proposition:
First, let me write down the theorem on which the subsequent proposition relies.
Theorem. (Euclid). There are infinitely many primes.
Proof: Suppose that there are finitely many primes, say, $p_1, p_2, \ldots, p_n$. Consider $$m=p_1 p_2 \cdots p_n$$ Consider $$m=p_1 p_2 \cdots p_n+1\ge 2$$
By the Fundamental Theorem of Arithmetics, $m$ is a product of primes. Thus $p_k \mid m$ for some $1\le k \le n$. Since $p_k\mid m$ and $p_k\mid p_1 p_2 \cdots p_n$, we have that $p_k \mid (m-p_1 p_2 \cdots p_n)$, i.e. $p_k\mid 1$, which leads to a contradiction. Thus there are infinitely many primes.
$$$$
Proposition: For $n\in\mathbb{N}$, we have $p_n\le 2^{2^n}$, where $p_n$ is the $n$-th prime number. The proof by induction goes as follows. For $k=1$, we have $p_1 = 2\le 2^2=4$. Suppose that the result holds for $1\le k \le n$ . We have seen in the proof of the Theorem above that
$$p_{n+1}\le p_1p_2\cdots p_n+1$$
Thus, by our induction hypothesis,
$$p_{n+1}\le 2^{2^1} 2^{2^2}\cdots 2^{2^n}+1=2^{2^{n+1}-2}+1\le 2^{2^{n+1}}$$
By induction, proposition holds.
What I don't seem to get is how it follows from the Theorem above that $$p_{n+1}\le p_1p_2\cdots p_n+1$$ Moreover, how can one immediately see (other than by proof by induction) that $2^1 + 2^2 + \cdots + 2^n = 2^{n+1}-2$?
Would appreciate some clarifications.
To see that $p_{n+1}\le p_1p_2\dots p_n+1$, note that $p_1p_2\dots p_n+1$ must have a prime factor, but it cannot be any of $p_1, p_2, \dots, p_n$ by the same argument as in the proof. So any prime factor of $p_1p_2\dots p_n+1$ must be a new prime number less than or equal to $p_1p_2\dots p_n+1$. Therefore, the smallest prime larger than $p_n$ can be at most $p_1p_2\dots p_n+1$.
As for why $2^1 + 2^2 + \dots + 2^n = 2^{n+1}-2$, write it in binary.