Suppose that the coefficients of the equation $x^n+a_{n-1}x^{n-1}+...+a_1x+a_0=0$ are real and satisfy $0<a_0 \leq a_1 \leq \dotsb \leq a_{n-1} \leq 1$. Let $z$ be a complex root of the equation with $\lvert z \rvert \geq 1$. Show that $z^{n+1}=1$.
This question seems very hard for me and I don’t know where to start.
Edit: From the post: $x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0=0$ has real coefficients which satisfy $0<a_0 \le a_1 \le \cdots \le a_{n-1} \le 1$ prove that $z$ is a root
I found the solution by Mr. Y, but he did not legally prove the question, and the only answer is too hard for me to understand.