I am surprised that no one mentioned this:

$$2 \cdot 2 \cdot 2... \cdot 2 \leq 2 \cdot 3 \cdot 4... \cdot (n-1)$$

Thus $2^{n-2} \leq (n-1)!$.

Hence we have

$$0 \leq \frac{2^n}{n!} \leq \frac{4(n-1)!}{n!}=\frac{4}{n} \,.$$

**Generalization**

Let $x$ be any real number.

Fix an integer $k$ so that $\left| x \right| <k$.

Then, for all $n> k$ we have:

$$\left| x\right| ^{n-k} < k(k+1)(k+2)...(n-1) $$

Thus

$$0 < \frac{\left|x \right|^n}{n!} \leq \frac{\left|x\right|^kk(k+1)(k+2)...(n-1)}{n!}=\frac{\left|x \right|^k}{(k-1)!}\frac{1}{n}$$

Since $k$ is fixed, $\frac{\left|x \right|^k}{(k-1)!}$ is just a constant, thus $\lim_n \frac{\left|x \right|^k}{(k-1)!}\frac{1}{n}=0$.

By Squeeze theorem, we get that

$$\lim_n \left| \frac{x ^n}{n!} \right|= \lim_n \frac{\left|x \right|^n}{n!}=0 \,.$$

Now, since $\lim_n \left| \frac{x ^n}{n!} \right|=0$, we get

$$\lim_n \frac{x ^n}{n!} = 0\,.$$

**P.S.** A more general result applicable in this case is the following:

**Lemma** If $a_n$ is a sequence so that

$$\limsup_n |\frac{a_{n+1}}{a_n}| <1$$
then
$\lim_n a_n =0$.