 Solution

As appropriate for a genuine $k$-step linear multistep method, take ${\alpha }_k\ne 0$. Suppose the characteristic polynomial has a polynomial root $\xi$ with modulus $|\xi |>1$. From $\rho (\xi )=0$ and the triangle inequality,Dividing by the positive number $|{\alpha }_k||\xi {|}^{k-1}$ gives $|\xi |\le 1$, a contradiction. HenceThe nonzero leading coefficient matters: if ${\alpha }_k=0$, the assumed inequality forces all the ${\alpha }_j$ to vanish, so $\rho$ is the zero polynomial rather than a characteristic polynomial with $k$ roots. This degenerate case does not define the method in the question.