I'm trying to solve an exercice:
Let $q$ be a root of the characteristic polynomial of the recurrence relation:
$$x_n=\alpha_1x_{n-1}+\alpha_2x_{n-2}+\ldots+\alpha_kx_{n-k}$$
If the multiplicity of the root $q$ is $m$, show that $x_n = n^iq^n$ is a solution of the recurrence relation above, for $i\in\{0, 1, \ldots, m − 1\}$.
I understand that $q^n$ is a solution of the linear recursion and that if $p(x)$ is the characteristic polynomial then $p^{(i)}(q)=0$ but I don't see how to use it to make a proof.
If the polynomial $$x^n-(\alpha_1x^{n-1}+\alpha_2x^{n-2}+\cdots\alpha_kx^{n-k})$$
has a root of degree $m$ at $x$, then all its $m$ derivatives also vanish at $x$, and
$$nx^{n-1}-(\alpha_1(n-2)x^{n-1}+\alpha_2(n-3)x^{n-2}+\cdots\alpha_k(n-k)x^{n-k-1})=0,$$
so
$$nx^{n}-(\alpha_1(n-1)x^{n-1}+\alpha_2(n-2)x^{n-2}+\cdots\alpha_k(n-k)x^{n-k})=0,$$
and you can iterate, because $(xp(x))'=p(x)+xp'(x)$.