How to find the limit of the sequence $a_n=\frac{n^n}{3^n\cdot n!}$ as $n$ tends to infinity. 
Let $a_n=\dfrac{n^n}{3^n\cdot n!}.$ Show that $a_n\to0$ as $n\to\infty$.

I know that I can use the ratio test for sequences, and $n^n$ increases faster than $3^n \cdot n!$ so it will tend towards infinity so I invert the sequence so I have to show that $\frac{3^n \cdot n!}{n^n}$ tends towards $0$.
I divide $a_{n+1}$ by $a_n$, I get $$\frac{3^{n+1} \cdot (n+1)! \cdot n^n}{(n+1)^{n+1} \cdot 3^n \cdot n!}.$$ I can get simplify up to get $\frac{3n^n}{(n+1)^{n+1}}$ but I do not know how to simplify any further so I can find the limit as $n$ tends to infinity.

A Third Proof
Since $$\frac{a_{n}}{a_{n-1}}=\frac{1}{3}\left(1+\frac{1}{n-1}\right)^{n-1}$$ for $n=2,3\cdots$
then $$a_{n}=a_1\cdot\prod_{k=2}^{k=n}\frac{a_{k}}{a_{k-1}}=a_1\cdot \prod_{k=2}^{k=n}\frac{1}{3}\left(1+\frac{1}{k-1}\right)^{k-1}.$$
Notice that $$\left(1+\frac{1}{k-1}\right)^{k-1}<e$$ for $k=2,3,\cdots$
Thus, $$0<a_{n}<\frac{1}{3} \cdot \left(\frac{e}{3}\right)^{n-1}$$for $n=2,3\cdots.$
Let $n \to \infty$. Since $0<\dfrac{e}{3}<1$,then $\left(\dfrac{e}{3}\right)^{n-1} \to 0$. By the squeeze theorem, we may claim $$\lim_{n \to \infty}a_n=0.$$
