Proof left as an exercise for the reader: For a non-trivial complex polynomial and any given input value $a$, there is a unique integer $k\geq1$ so that $$p(z)-p(a)=(z-a)^kq_k(z), q_k(a)\neq0.$$
My intuition here is to use something similar to proof by induction and Taylor expansion, but I'm not sure if this would be a proof of the exercise.
Sketch of induction proof. By the Factor Theorem we have $$p(z)-p(a)=(z-a)q(z)$$ for some $q(z)$. If $q(a)\ne0$ we are finished. If $q(a)=0$ we can use induction, since $q$ has smaller degree than $p$, and we get $$q(z)=q(z)-q(a)=(z-a)^kr(z)\ ,\qquad r(a)\ne0\ .$$ Substituting back completes the proof.
Will leave you to supply details :)
For uniqueness, if $$(z-a)^kq_k(z)=(z-a)^lq_l(z)\ ,\quad q_k(a)\ne0\ ,\quad q_l(a)\ne0$$ with $k\ne l$, then we can assume $k>l$, so $z-a$ is a factor of $q_l(z)$. But this contradicts $q_l(a)\ne0$.