Use the well known theorem that 2< \left(1+\dfrac{1}{n} \right)^n<3
Let a_n = \left( \dfrac{n}{3}\right)^n; b_n=n!, c_n = \left( \dfrac{n}{2}\right)^n
\dfrac{a_{n+1}}{a_n} = \left(\dfrac{n+1}{n}\right)^n \times \dfrac{n+1}{3}<n+1 = \dfrac{b_{n+1}}{b_n}
i.e. the sequence a_n grows slower than b_n and clearly a_1<b_1. Hence a_n<b_n \ \forall \ n \in \mathbb{N}
