Euclid's Theorem/Corollary 2/Proof 2

Proof
there exists a largest prime number $p$.

Let $b = p! + 1$.

Let $q$ be a prime number that divides $b$.

Since $p$ is the largest prime number, $q \le p$.

However, no positive integer $d \le p$ is a divisor of $b$.

Hence $q \not \le p$.

Hence the result, by Proof by Contradiction.


 * : $\S 1.12$: Valid Arguments: Proposition $1.12.2$