Both directions follow from the general fact that, in a nonempty linear order with a locally finite bottom, a set is infinite iff for every \(a\) some element of the set exceeds \(a\) (Set.infinite_iff_exists_gt). Specializing to \(\mathbb {N}\) and the set \(\{ p : \mathbb {N} \mid p \text{ is prime}\} \) gives the claim.

Suppose for contradiction that the set of primes is finite. Then it is bounded above by some \(n \in \mathbb {N}\), i.e., every prime \(p\) satisfies \(p \le n\). Consider \(N := n! + 1\); since \(N \ge 2\), it has a prime divisor \(p\). Because \(p\) is prime and \(p \le n\), we have \(p \mid n!\), hence \(p \mid N - n! = 1\), contradicting that \(p\) is prime.

Suppose for contradiction that the set of primes \(S\) is finite. Consider \(N := \prod _{p \in S} p + 1\). Since \(N {\gt} 1\), it has a prime divisor \(p\). But then \(p \in S\), so \(p \mid \prod _{p \in S} p\), hence \(p \mid N - \prod _{p \in S} p = 1\), contradicting that \(p\) is prime.

Since \(2^{2^n} \ge 1\), we have \(F_n = 2^{2^n} + 1 \ge 2\).

Since \(2^n \ge 1\), \(2^{2^n}\) is even, so \(F_n = 2^{2^n} + 1\) is odd.

We prove by induction that \(\prod _{k=0}^{n-1} F_k = F_n - 2\). The base case \(n = 0\) is trivial since the empty product is \(1 = 3 - 2 = F_0 - 2\). For the inductive step, assuming \(\prod _{k{\lt}n} F_k = F_n - 2\): \(\prod _{k{\lt}n+1} F_k = (F_n - 2) F_n = (2^{2^n} + 1)(2^{2^n} - 1) = 2^{2^{n+1}} - 1 = F_{n+1} - 2\). Adding \(2\) to both sides and using \(F_{n+1} \ge 2\) gives the claim.

By Lemma lemma 18, \(F_{n+1} = \prod _{k=0}^{n} F_k + 2\), so \(\gcd (F_{n+1}, \prod _{k=0}^{n} F_k) = \gcd (2, \prod _{k=0}^{n} F_k)\). Since each \(F_k\) is odd by Lemma lemma 17, the product is odd, hence the gcd is \(1\).

WLOG \(m {\lt} n\). Write \(n = k + 1\). Then \(m \in \{ 0, \dots , k\} \), so \(F_m \mid \prod _{k'=0}^{k} F_{k'}\). By Lemma lemma 19, \(\gcd (F_n, \prod F_{k'}) = 1\), so \(\gcd (F_n, F_m) = 1\).

For each \(n \in \mathbb {N}\), \(F_n \ge 2\) has a smallest prime factor \(p_n\). By Lemma lemma 20, the Fermat numbers are pairwise coprime, so the map \(n \mapsto p_n\) is injective. Hence there are infinitely many primes.

Use the inequality \(\log (n + 1) \le H_n\).

Use the geometric expansion of \(p / (p - 1) = (1 - 1/p)^{-1}\) for each prime \(p\). Then the product is equal to the sum of inverses of the integers whose primes factors are all at most \(n\), and the sum is greater than or equal to the n-th harmonic number.

Assume that there are only finitely many primes. Then the product of \((1 - 1/p)^{-1}\) over all primes \(p\) is finite. But by theorem 23, this product is greater than the harmonic series, which is unbounded (theorem 22), a contradiction.

By induction. The base case is \(a_0 = 2\). For the inductive step, if \(a_n \ge 2\) then \(a_{n+1} = a_n (a_n + 1) \ge 2 \cdot 1 = 2\).

By induction on \(n\). The base case is that \(a_0 = 2\) has the prime divisor \(2\). For the inductive step, since \(a_n\) and \(a_n + 1\) are consecutive they are coprime, so their prime factors are disjoint and the prime factors of \(a_{n+1} = a_n (a_n + 1)\) split as a disjoint union. By induction \(a_n\) has at least \(n + 1\) prime divisors, and since \(a_n + 1 \ge 3\) it has at least one prime divisor. So \(a_{n+1}\) has at least \(n + 2\) distinct prime divisors.

By lemma 27, for every \(n\) the integer \(a_n\) has at least \(n + 1\) distinct prime divisors. If the set of primes were finite, then for every \(n\) these prime divisors would be a subset of the finite set of all primes, bounding \(n + 1\) above by a fixed constant — a contradiction.

By direct computation.

Since \(p \ge 3\) and \(F\) is monotone, \(F_p \ge F_3 = 2\).

Distinct primes are coprime, so \(\gcd (p, q) = 1\). Since \(\gcd (F_m, F_n) = F_{\gcd (m, n)}\), we have \(\gcd (F_p, F_q) = F_1 = 1\).

Assume that there are only finitely many prime numbers, \(\{ p_1 = 2, p_2 = 3, \ldots , p_n\} \). Consider the corresponding Fibonacci numbers \(F_{p_2}, \ldots , F_{p_n}\) (except for \(F_{p_1} = F_2 = 1\)). By \(\gcd (F_m, F_n) = F_{\gcd (m, n)}\), these are pairwise coprime. Since we have \(n\) primes and \(n - 1\) Fibonacci numbers, each of the Fibonacci numbers can have at most two prime factors. However, this is not possible since \(F_{37} = 73 \cdot 149 \cdot 2221\) is a product of three distinct primes.