Answer
See explanation
Work Step by Step
**Proof:**
Suppose, for the sake of contradiction, that there are only finitely many prime numbers. Then there is a largest prime; call it \( p \).
Define
\[
M = p! + 1.
\]
Since \( p! \) (the factorial of \( p \)) is divisible by every prime number less than or equal to \( p \), for any prime \( q \) with \( q \le p \) we have
\[
q \mid p!.
\]
Thus, when dividing \( M \) by any prime \( q \le p \),
\[
M = p! + 1 \equiv 0 + 1 \equiv 1 \pmod{q}.
\]
That is, none of the primes \( \le p \) divides \( M \).
Now, by the Fundamental Theorem of Arithmetic, every integer \( M > 1 \) has at least one prime divisor. Let \( q \) be any prime divisor of \( M \). Since no prime less than or equal to \( p \) divides \( M \), it must be that
\[
q > p.
\]
This contradicts our assumption that \( p \) is the largest prime. Therefore, there must be infinitely many prime numbers.
