Answer
See explanation
Work Step by Step
**Proof:**
Let \( n \) be an integer with \( n > 2 \). Consider the number
\[
M = n! - 1.
\]
By the Fundamental Theorem of Arithmetic, \( M \) has at least one prime divisor. Let \( p \) be a prime number such that \( p \mid M \).
**Claim:** \( p > n \).
**Justification:**
Assume for contradiction that \( p \le n \). Then, since
\[
n! = 1 \cdot 2 \cdot 3 \cdots n,
\]
\( p \) divides \( n! \). Therefore, \( p \) divides both \( n! \) and \( n! - 1 \). Consequently, \( p \) must divide their difference:
\[
n! - (n! - 1) = 1.
\]
However, no prime number divides 1. This contradiction shows that our assumption is false; hence, \( p > n \).
Also, since \( p \) is a divisor of \( M = n! - 1 \), it follows that
\[
p \le n! - 1 < n!.
\]
Thus, we have
\[
n < p < n!.
\]
Since \( p \) is a prime number with \( n < p < n! \), this completes the proof.
