A simple dynamic programming problem.
Define array dp, dp[i] means the answer when n === i, if you want to get dp[10], you may compare dp[1] * 9, dp[2] * 8, dp[3] * 7 ... That's really dynamic programming by O(n^2) Complexity.
To make it easier, I consider the case n === 2 and n === 3 as the special cases to avoid some error. That's because dp[i] may be made up by only one factor, dp[2] is 2, not 1, and dp[3] is 3, not 2, after dp[3], we can get correct answer as we expect.