Squeeze theorem to prove that the sequence $a_n= \frac{3^n}{n!}$ is convergent? I want to know if the following proof of the convergence of a sequence is correct.
Proof. Let $a_n= \frac{3^n}{n!}$. Firstly, it is trivial to see that $a_n \geq 0$ for all $n \in \mathbb{N}$. Secondly, see that
$$a_n = \frac{3^n}{n!}= \frac{3 \times 3\times3 \times \cdot\cdot\cdot \times 3 \times 3 \times 3}{n \times (n-1) \times (n-2) \times \cdot \cdot \cdot 3 \times 2 \times 1} = \frac{3\times3\times3\times3}{n\times3\times2\times1} \times \frac{3 \times 3 \times...\times 3}{(n-1) (n-2)\cdot\cdot\cdot \times 4}$$
$$= \frac{3^4}{6n} \times \Big(\frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4}\Big)$$
Because $0 < \frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4} <1$ we know $a_n=\frac{3^4}{6n} \times \Big(\frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4}\Big) \leq \frac{3^4}{6n}$. Using the fact that $a_n \geq 0$ we conclude that, since
$$0 \leq a_n \leq \frac{3^4}{6n}$$
and $0 \to 0, \frac{3^4}{6n} \to 0$ when $n \to 0$,
$$\lim_{n\to\infty}a_n=0$$
I skipped some trivial steps (e.g., showing that $\frac{3^4}{n}$ tends to $0$ if $n \to \infty$) because I assume theorems that prove such properties. I am mostly concerned about whether my manipulation of the factorial expression is correct and if the squeeze theorem is properly applied.
Thanks in advance!

Observe that: $0<a_{n+1} = \dfrac{3}{n+1}a_n < \dfrac{3}{n+1}\cdot \dfrac{3}{2}= \dfrac{9}{2n+2}$ for $n \ge 2$. Thus the squeeze theorem can be applied to conclude that $a_{n+1} \to 0$  as claimed.
