Does this problem have to do with something like? $\lambda^{n+1}=1$ $\lambda^{n+1}-1=0$ $\frac{\lambda^{n+1}-1}{\lambda-1}=0$

$\lambda^n=-(a_{n-1}\lambda^{n-1}+a_{n-2}\lambda^{n-2}+\cdots+a_0)$ $\lambda^{n+1}=-(a_{n-1}\lambda^n+a_{n-2}\lambda^{n-1}+\cdots+a_0\lambda)$ $\lambda^{n+1}-\lambda^n=-a_{n-1}\lambda^n+(a_{n-1}-a_{n-2})\lambda^{n-1}+\cdots+(a_1-a_0)\lambda+a_0$ $\implies |\lambda|^{n+1}=|(1-a_{n-1})\lambda^n+(a_{n-1}-a_{n-2})\lambda^{n-1}+\cdots+(a_1-a_0)\lambda+a_0|$ $\le (1-a_{n-1})|\lambda|^n+(a_{n-1}-a_{n-2})|\lambda|^{n-1}+\cdots+(a_1-a_0)|\lambda|+a_0$ $\le (1-a_{n-1})|\lambda|^n+(a_{n-1}-a_{n-2})|\lambda|^n+\cdots+(a_1-a_0)|\lambda|^n+a_0|\lambda|^n = |\lambda|^n$ $\implies |\lambda| \le 1 \implies |\lambda|=1$. Thus, the triangle inequality holds and $a_0=a_1=\cdots=a_{n-1}=1 \implies \lambda^{n+1}=1$

Shouldn't it be $|\lambda| \geq 1$ in the question?