calculus - Alternative way to prove $lim_{ntoinfty}frac{2^n}{n!}=0$?

It follows easily from the convergence of $\sum_{n=0}^\infty\frac{2^n}{n!}$ that

$$
\lim_{n\to\infty}\frac{2^n}{n!}=0\tag{1}
$$
Other than the Stirling's formula, are there any "easy" alternatives to show (1)?

Answer

Yes: note that
$$ 0\leq \frac{2^n}{n!}\leq 2\Big(\frac{2}{3}\Big)^{n-2}$$
for $n\geq 3$, and then use the squeeze theorem.