Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$.
Let $a=z^{120}$. This simplifies the problem constraint to $a^6-a \in \mathbb{R}$. This is true if $\text{Im}(a^6)=\text{Im}(a)$. Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$. We are given $\sin\theta = \sin{6\theta}$. Note that $\sin \theta = \sin (\pi - \theta)$ and $\sin \theta = \sin (\theta + 2\pi)$. We can use these facts to create two types of solutions: 
\[\sin \theta = \sin ((2m + 1)\pi - \theta)\]
which implies that $(2m+1)\pi-\theta = 6\theta$ and reduces to $\frac{(2m + 1)\pi}{7} = \theta$. There are 7 solutions for this.
\[\sin \theta = \sin (2n\pi + \theta)\]
which implies that $2n\pi+\theta=6\theta$ and reduces to $\frac{2n\pi}{5} = \theta$. There are 5 solutions for this, totaling 12 values of $a$.
For each of these solutions for $a$, there are necessarily $120$ solutions for $z$. Thus, there are $12\cdot 120=1440$ solutions for $z$, yielding an answer of $\boxed{440}$.