Yes, a function $f$ is said to be negligible if for every polynomial function $p(n)$ there exits some constant $N$ such that $f(n) < \frac{1}{p(n)}$ for all $n > N$.
If $\frac{1}{n!} < \frac{1}{p(n)}$ then $n! > p(n)$, for all polynomials $p(n)$ and suitable $N$ such that $n>N$. Thus, you'd only have to prove the second segment of the statment.
The EASIEST solution is to note that $n!$ can be approximated by Stirling's Approximation as $$\sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n$$ and so $$\frac{1}{n!} = \frac{1}{\sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n}$$ which is clearly an exponentially decaying function and therefore will always be less than any polynomial function $p(n)$ for such that $n>N$ for suitable $N$.