Supremum and infimum of set with n!

Question

I want to conclude about $\sup A$ and $\inf A$ where $A = \{ {n^n \over (n!)^2}: n = 1, 2, \dots \}$. My intuition is that $\sup A = 1$ and $\inf A=0$. To show that the infimum exists, I could prove that ${n^n \over (n!)^2}$ can be "as small as we wish" and compare it with $\inf A + \epsilon$ for whatever $\epsilon$ we choose. The problem here is that limits haven't been properly introduced in the material I'm going over. How can I go about this?