Let $V$ be a vector space and $f \in \text{End}V$. Let $p$ be a polynomial over a field $K$ so that $p(f)=0$. Also we have deg $p = m$ and $c_1,c_2,\dots,c_m$ be all roots of $p$. Prove that at least one $c_i$ is an eigenvalue of $f$.
So if $c_i$ are all roots we can conclude that $$ p(x) = k(x-c_1)\dots(x-c_m) $$ If we substitute $f$ for $x$ then we get $$ p(f) = k(f - c_1)\circ\dots\circ(f-c_m) = 0 $$ How to proceed from here?
You have $p(x)=k(x-c_1)^{\alpha_1}\cdots(x-c_m)^{\alpha_m}$ for some $k\in K\setminus\{0\}$ and some natural $\alpha_j$'s (you can't conclude that the $\alpha_j$'s are equal to $1$). And$$p(f)=0\iff k(f-c_1\operatorname{Id})^{\alpha_1}\circ\cdots\circ(f-c_m\operatorname{Id})^{\alpha_m}=0.$$But then some $f-c_k\operatorname{Id}$ is not invertible and therefore $c_k$ is an eigenvalue of $f$.