real analysis - Proof that limit as n approaches inf of $3^n cdot frac{1}{n!} = 0$ using squeeze theorem

This is a homework question. I'm completely new to epsilon proofs, so I'm pretty bad at them.

I want to prove that $3^n \cdot \frac{1}{(n!)} = 0$ converges to 0.

Here's where I'm at. By the definition of a limit of a sequence, I want

$\forall \epsilon>0 \exists N \in \mathbb{N}$ such that $\mid \frac{3^n}{n!} - 0 \mid < \epsilon$.

Or, I want to use the squeeze theorem to bound it below and above with functions with limits of 0.

I started to try to get an appropriate value for N, but it doesn't seem to work out... I'm using the arbitrarily small $\epsilon$ to try to find an N for which $\frac{3^n}{n!} < \epsilon$ for all n > N. I've just tried doing algebraic manipulation, but I haven't got anywhere.

For the squeeze theorem, I thought of trying to bound the sequence between the sequences $x_n = \frac{1}{n}$ and $z_n = 0$. But I would have to prove that $\frac{3^n}{n!}$ is bounded above by $\frac{1}{n}$, which I'm not sure I know how to do.

If you could tell me which of the two methods would be appropriate and some hints for how to proceed I would appreciate it, thanks.