Answer
Eventually strictly increasing.
Work Step by Step
We have: $a_n=\dfrac{n !}{3^n}$ and $a_{n+1} =\dfrac{(n+1)!}{3^{n+1}}$
Then
$\dfrac{a_{n+1}}{a_n}=\dfrac{\dfrac{(n+1)!}{3^{n+1}}}{\dfrac{n !}{3^n}}\\=\dfrac{ 3^n (n+1)!}{3^{n+1}n!}\\=\dfrac{n+1}{3} \gt 1 \ for \ n \gt 2$
This implies that the given sequence is eventually strictly increasing.