Specialize the general characterization of infinite subsets of a locally finite linear order to the set of natural primes.

If the set of primes were finite, a prime divisor of one plus their product would already lie in the set, and therefore divide both the product and the product plus one.

If all primes were bounded by n, a prime divisor of n! + 1 would both divide n! and divide n! + 1, hence divide 1, impossible; so the primes are infinite Definition 2.1.

Since 2^{2^n} \ge 1, adding one gives F_n \ge 2.

The exponent 2^n is positive, so 2^{2^n} is even and 2^{2^n}+1 is odd.

Inductively prove that \prod_{k<n} F_k = F_n - 2, then rewrite the next product using the difference-of-squares identity.

The recurrence writes F_{n+1} as the product plus 2. The product is odd, so the only possible common divisor with 2 is 1.

For m<n, the number F_m divides the product of the earlier Fermat numbers, which is coprime to F_n.

Each Fermat number has a prime divisor. Pairwise coprimality Lemma 2.11 makes the chosen prime divisors pairwise distinct, producing infinitely many primes.

Use the inequality \log(n+1) \le H_n and the fact that the logarithm is unbounded.

Expand each factor p/(p-1) as a finite geometric sum. The resulting product contains terms corresponding to reciprocals of positive integers up to n.

If there were only finitely many primes, the Euler product over all of them would be fixed. But the product over primes below a sufficiently large bound Theorem 2.14 dominates arbitrarily large harmonic sums Theorem 2.13, a contradiction.

The base case is 2; the inductive step multiplies two positive factors and stays at least 2.

Consecutive numbers a_n and a_n+1 are coprime, so the prime factors of a_{n+1} split as a disjoint union of the prime factors of the two factors.

If only finitely many primes existed, the number of prime factors of any a_n would be bounded by that finite set, contradicting the previous lemma Lemma 2.18 for large n.

This is a direct computation.

An odd prime is at least 3, and the Fibonacci sequence is monotone, so F_p \ge F_3 = 2.

Distinct primes are coprime, and \gcd(F_m,F_n) = F_{\gcd(m,n)} reduces the result to F_1=1.

Assuming finitely many primes, look at Fibonacci numbers indexed by odd primes, each at least 2 Lemma 2.21. They are pairwise coprime Lemma 2.22, but F_{37} already has three distinct prime factors Lemma 2.20, contradicting the finite counting bound.

Assume that there are only finitely many such primes. Let P be the product of all such primes and consider N = 4P^2 + 1. If we choose a prime factor q of N, then (2P)^2 \equiv -1 \pmod{q}. Hence -1 is a square modulo q, so q \equiv 1 \pmod{4}, which contradicts the assumption that P is the product of all such primes.

This follows from Theorem 2.24.

If every prime factor of n were congruent to 1 modulo 4, then n would be congruent to 1 modulo 4, contradicting the assumption.

Assume that there are only finitely many such primes. Let P be the product of all such primes and consider N = 4P - 1. Since N \equiv 3 \pmod{4}, it has a prime factor q with q \equiv 3 \pmod{4} Lemma 2.26, which contradicts the assumption that P is the product of all such primes.

This follows from Theorem 2.27.

Set I_n = \int_0^1 (x-x^2)^n \sin(\pi x)\,dx. Since (x-x^2)^n vanishes at both endpoints, integrating by parts twice gives \pi^2 I_{n+2} = 2(n+2)(2n+3) I_{n+1} - (n+2)(n+1) I_n. Suppose \pi^2 = a/b with a, b integers and b > 0. The recursion then shows that the integers defined by A_0 = 2, A_1 = 4b and A_{n+2} = 2(2n+3) b A_{n+1} - ab A_n satisfy A_n \cdot n! = b^n \pi^{2n+1} I_n. This is Niven's assertion that \pi \int_0^1 a^n x^n (1-x)^n \sin(\pi x)/n!\,dx is an integer, which he reads off from the auxiliary function g = b^n \sum_k (-1)^k \pi^{2n-2k} f^{(2k)} for f(x) = x^n(1-x)^n/n!; the telescoping property of g is exactly the recursion above. The integrand of I_n is positive on (0,1), so each A_n is a positive integer and A_n \ge 1. But 0 \le x - x^2 \le 1/4 and \sin(\pi x) \le 1 on [0,1] give I_n \le 4^{-n}, hence A_n \le \pi (a/4)^n / n!, which tends to 0. Contradiction.

If there were only finitely many primes, the partial Euler products \prod_{p < n} (1 - p^{-2})^{-1} would be constant for large n, so Euler's product formula \zeta(2) = \prod_p (1 - p^{-2})^{-1} would exhibit \zeta(2) as a finite product of rational numbers, hence as a rational number. But \zeta(2) = \pi^2/6 and \pi^2 is irrational Theorem 2.29, a contradiction.