Since $(-4)^n$ alternates between negative and positive, an obvious "sandwiching" will be between $\frac{4^n}{n!}$ and $-\frac{4^n}{n!}$. One way to prove that $\frac{4^n}{n!}$ goes to 0 is to show that $\sum \frac{4^n}{n!}$ converges since a necessary condition that a series converge is that the sequence of term go to 0. You can do that by using the "ratio test" or even by noting that the series converges to $e^4$.