Suppose there are finitely many primes \(\{p_1,p_2, \ldots, p_n \}\), with \(n ≥ 1\). Consider \(N\):
$$N = (p_1p_2 \ldots p_n) + 1$$
\(N\) is either a prime or not. If it's a prime, then there is a contradiction, if not, then by the Fundamental Theorem of Arithmetic there must be a prime \(p\) dividing N. If this \(p\) exists in \( (p_1p_2 \ldots p_n) \), then \(p\) divides both \( (p_1p_2 \ldots p_n) \) and \(N\). If \(p\) divides both, then it would also divide the difference (which is 1):
$$\begin{gather} p | (p_1p_2 \ldots p_n) \wedge p | N \implies \\ p | (p_1p_2 \ldots p_n) - N \\ p | 1 \end{gather} $$
Since \(p\) dividing 1 is not possible, \(p\) is a different prime not in \((p_1p_2 \ldots p_n)\), which is also a contradiction. Therefore, there are not finitely many primes.