Let $z = e^{2\pi i\alpha}$ with $\alpha \in [0, 1[$ irrational. I'm convinced the sequence $(z^{N!})_{N \in \mathbb{N}}$ has no reason to ever stabilize but I'm not sure how to prove or disprove this.
If we suppose it converges to 1 for example, then we know there exists $M \in \mathbb{N}$ such that $N \ge M \implies d(z^{N!}, 1) < \frac{1}{2}$ and at the same time, the irrationality of $\alpha$ implies that there exists an infinite amount of natural numbers r such that $d(z^r, 1) \ge \frac{1}{2}$. This does seem like an absurd thing, but the problem is nothing forces those r's to be factorials, and so it could just happen that these two things don't contradict each other.
I'd appreciate any help.
It can stabilize, and in fact a handy example is $\alpha=e$!
Since $\displaystyle e=\sum_{k=0}^\infty \frac1{k!}$, we have $\displaystyle N! e = \sum_{k=0}^\infty \frac{N!}{k!} = M_N + \beta_N$ where $\displaystyle M_N = \sum_{k=0}^N \frac{N!}{k!}$ is an integer and $\displaystyle 0< \beta_N = \sum_{k=N+1}^\infty \frac{N!}{k!} < \frac2{N+1}$. And since $z^{N!} = e^{i(2\pi e N!)} = e^{2\pi i(M_N+\beta_N)} = e^{2\pi i\beta_N}$, we conclude that $\lim\limits_{N\to\infty} z^N = 1$ when $z=e^{2\pi ie}$.