Consider recursion $$a_n+c_1a_{n-1}+\cdots+c_ma_{n-m}=0~~~~~~~~~~(1)$$
Let $\lambda^m+c_1\lambda^{m-1}+\cdots+c_m=0$ be the characteristic function and $(\lambda_1,n_1),\cdots,(\lambda_s,n_s)$ be its solutions (the $\lambda_i$ are the roots with multiplictiy $n_i$, so that $n_1+\cdots+n_s=m$).
Now I want to prove that
$\lambda_1^n,~n\lambda_1^n,~\cdots,~n^{n_1-1}\lambda_1^n$
$\lambda_2^n,~n\lambda_2^n,~\cdots,~n^{n_2-1}\lambda_2^n$
$\cdots\cdots$
$\lambda_s^n~,n\lambda_s^n,~\cdots,~n^{n_s-1}\lambda_s^n$
are $m$ independent linear solutions of $(1)$.
Below is my outline of proof.
First of all, they are solutions which I have done.
Then they must be independent. My thought is to focus on the next question.
Let $P_1(x),\cdots P_s(x)\in\mathbb C[x]$ and $\lambda_1,\cdots\lambda_s\in\mathbb C$. If $$P_1(n)\lambda_1^n+\cdots+P_s(n)\lambda_s^n=0~~~n\in\mathbb N$$ then $P_1(x)=\cdots=P_s(x)=0$.
I get stuck here. Some advice on proof of the question. Thank you.