Let $p: \mathbb{F} \to \mathbb{F}$ be a polynomial of degree of at most $m$, i.e. $p \in \mathcal{P}_m(\mathbb{F})$ such that $p(r) = 0$ for some $r \in \mathbb{F}$. Is it true that in that case $p$ can always be written as $$p(z) = b_1(z- r) + b_2(z-r)^2 + \cdots + b_m(z-r)^m$$ for some $b_i \in \mathbb{F}$, i.e. $p \in \mathrm{span}(z-r, (z-r)^2, \dots (z-r)^m)$?
I suspect it could. But maybe I'm wrong.
My proof attempt: Let $p \in \mathcal{P}_m(\mathbb{F})$ be such that $p(r) = 0$. Then $p(z) = a_0 + a_1z + \cdots + a_mz^m$ and $a_0 = -a_1r - \cdots - a_mr^m$. We can find the coefficients $b_i$ by solving the system of linear equations ($f_i$ are linear functions): $$ \begin{cases} a_0 = f_0(b_1, \dots, b_m)\\ a_1 = f_1(b_1, \dots, b_m)\\ a_2 = f_2(b_2, \dots, b_m)\\ \vdots \\ a_m = f_m(b_m) \end{cases} $$ From the last equation we get $b_m$, from the second to last we get $b_{m-1}$, etc. But there are two equations that can be used to obtain $b_1$ which could lead to a contradiction. But $f_0$ is a constant function ($a_0$ depends only on $a_1, \dots, a_m$ and $r$). So this system always allows us to find unique $b_i$.
Is this proof correct?