For a positive integer \(n\), the product of the integers from \(1\) to \(n\) can be written in abbreviated form as \(n!\), which we read as “\(n\) factorial”. So, \[n! = n \times (n-1)\times(n-2)\times \cdots \times 3 \times 2\times 1\]
For example,
\(6! = 6 \times 5 \times 4\times 3\times
2\times 1 = 720\), and \(11! = 11\times
10 \times 9 \times \cdots \times 3 \times 2 \times 1 =
39\,916\,800\).
Note that \(6!\) ends in one zero and \(11!\) ends in two zeros.
Determine the smallest positive integer \(n\) such that \(n!\) ends in exactly \(1000\) zeros.