Characteristic polynomial of left multiplication equals minimal polynomial of element

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I'm trying to answer the following question:

Let $E/F$ be an extension field of prime degree $l$ and let $α ∈ E \setminus F$. Let $M_α$ be $F$-linear map induced by the multiplication by $α$:

$M_α : E \rightarrow E \\ \space\space\space\space u \mapsto α · u$.

Show that the characteristic polynomial of $M_α$ is equal to the minimal polynomial of $α$.

I'm given a hint to use Cayley-Hamilton, and I can see that since $\chi(x) = \det (x I - M_α)$ we get $\chi(α) = 0$ as $(α I - M_α)(x) = αx - αx = 0 $ $\forall x$, and so applying Cayley-Hamilton we see that the minimal polynomial of $α$ divides the characteristic polynomial of $M_α$.

I can't, however, see how to go about showing the converse - I don't know how to show the characteristic polynomial divides the minimal and hence that they're equal.

If anyone can spot what I'm missing or offer another solution I'd really appreciate it!