Hint: $\dfrac{n^n}{n!} \ge n$.
Added: Let $n \ge 2$. The number $n!$ is the product of the $n-1$ terms $n, n-1, n-2, \dots, 2$, all of which are $\le n$. Thus $n! \le n^{n-1}$, and therefore $\dfrac{n^n}{n!} \ge n$.
Now $a_n=n!\left(1-\frac{n^n}{n!}\right) \le 1-n.$ But $\displaystyle \lim_{n\to \infty}(1-n) =-\infty$. Since $a_n\le 1-n$, it follows that $\displaystyle \lim_{n\to \infty}(n!-n^n) =-\infty$.
Comment: In situations of this type, some people prefer to say that the limit does not exist. That assertion is less informative, since it tells us much less about the behaviour of $a_n$ as $n$ gets very large.