Assume the number of prime numbers is finite, with a total of $n$ primes, which are $p_{1}, p_{2}$, $p_{3}, \cdots, p_{n}$. Among them, $p_{1}=2, p_{2}=3, p_{3}=5, \cdots$. Let $a=p_{1} \cdots p_{n}+1$. If $a$ is a prime number, then because $a$ is not equal to any of $p_{1}, p_{2}, \cdots, p_{n}$, the number of prime numbers is at least $n+1$, which contradicts the assumption that there are only $n$ prime numbers. If $a$ is not a prime number, then by Lemma 5, the smallest factor $b$ of $a$ greater than 1 is a prime number. Since $p_{1} \cdots p_{n}$ is divisible by any of $p_{1}, p_{2}, \cdots, p_{n}$, but 1 is not divisible by any of $p_{1}, p_{2}, \cdots, p_{n}$, $a$ is not divisible by any of $p_{1}, p_{2}, \cdots, p_{n}$. Therefore, $b$ is not equal to any of $p_{1}, \cdots, p_{n}$, so there are prime numbers outside of $p_{1}, \cdots, p_{n}$.