Suppose that the coefficients of the equation $x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0=0$ are real and satisfy $0<a_0 \le a_1 \le \cdots \le a_{n-1} \le 1$.
Let $z$ be a complex root of the equation with $|z| \ge 1$. Show that $z^{n+1}=1$ $\space\space\space\space\space\space\space\space\space\space\space\space$(Source:MOP)
My work
Let $0<a_0=a_1 = \cdots = a_{n-1}=1 $, then, given that $z$ is a complex root of the equation $x^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0=0$, we have that $$z^n+z^{n-1}+\cdots+z+1=0 $$
Now,if we let $n=k-1$ we have $$z^{k-1}+z^{k-2}+\cdots+1=0 $$
From this it follows that $z$ is a root of unity which satisfy the equation $x^k-1=0$ ,so we have that $z^k=1$ where $k=n+1$ so we have $z^{n+1}=1$
That's what I've been able to do but I am still unsatisfied as I have considered a special condition, namely $0<a_0=a_1 = \cdots = a_{n-1}=1 $, and I would like to see how to take care also of the other condition.
How can I do that?
Thanks in advance
Edit: The current answer is just too hard for me to understand.
Consider $$(1-x)\left(x^n+\sum_{i=0}^{n-1} a_ix^i \right) = a_0 - x^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})x^i\qquad (1)$$ Let $x$ with $|x|>1$ be a root of $(1).$ Therefore, $x^{n+1}=a_0 +\sum_{i=1}^{n-1} (a_i-a_{i-1})x^i $ and we have with $a_n=1,$ \begin{align} | x^{n+1}| &= \left|a_0 + \sum_{i=1}^{n} (a_i-a_{i-1})x^i\right| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|x^i| \\ & <|x|^n\left( a_0 + \sum_{i=1}^n (a_i-a_{i-1})\right) \\ & = |x|^n\end{align} a contradiction.
Thus if there is a root such that $|x|\geq 1$ therefore $|x|=1.$