Product of any r consecutive natural numbers is always divisible by

For each n $$\in$$ N, 2³ⁿ $$-$$ 1 is divisible by

here N is a set of natural numbers.

The number (101)¹⁰⁰ $$-$$ 1 is divisible by

Let $$P(n) = {3^{2n + 1}} + {2^{n + 2}}$$ where $$n \in N$$. Then