What is the smallest positive integer n such that x^n - 1 is factored into precisely five irreducible factors?

Question

What is the least positive integer n that can be placed in the following expression:

n!(n+1)!(2n+1)! - 1

and yields a number ending with thirty digits of 9's.

Answer 1

n=16

Explanation

x^16 - 1
= (x^8+1)(x^8-1)
= (x^8+1)(x^4+1)(x^4-1)
= (x^8+1)(x^4+1)(x^2+1)(x^2-1)
= (x^8+1)(x^4+1)(x^2+1)(x-1)(x+1)