Suppose, for the sake of contradiction, that there is a finite number of prime numbers \((\{P_1, P_2, \dots, P_n\})\). Let us construct the integer \(x\) as follows:

$$x = (P_1 \times P_2 \times \dots \times P_n) + 1$$

By the Fundamental Theorem of Arithmetic, \(x\) must have at least one prime factor, let’s call it \(q\).

If \(q\) were one of the primes in our original set \(\{P_1, \dots, P_n\}\), then \(q\) would divide the product \((P_1 \dots P_n)\). Since \(q\) also divides \(x\), it must divide the difference:

$$x – (P_1 \times \dots \times P_n) = 1$$

However, no prime number divides 1. Therefore, \(q\) cannot be any of the primes \((P_1, \dots, P_n)\).

This means \(q\) is a prime number not included in our “finite” set. This contradicts the assumption that our set contained all prime numbers. Thus, the number of primes must be infinite. ■