Given a positive integer \(n\), consider the sequences of integers of the form \((n+1)!+i\) where \(i\) is an integer and \(2 \le i \le n+1\):
$$ (n+1)!+2, \ (n+1)!+3, \ \ldots, \ (n+1)! + (n+1) $$
The integer \(i\) (where \(2 \le i \le n+1\)) will have a duplicate inside \((n+1)!\), this means \(i \mid (n+1)! \).
If \(i \mid (n+1)! \) and \(i \mid i \), then \([ i \mid (n+1)! + i ]\) for all \(i \ge 2\) and \( i \le n+1\).
So each term in the sequence is divisible by \(i\), making them composite. In other words, there are \(n\) consecutive composite integers for a given \(n\).