Lemma: $End_k(V) \otimes k[t] \cong End_{k[t]}(V \otimes k[t])$ for $V$ a finite dimensional free $k$module (see below)
Corollary [The CayleyHamilton Theorem]: For an endomorphism $\phi : V \rightarrow V$ of a finite dimensional vector space $V$ over a field $k$, $p(\phi) = 0$ where $p \in End(V) \otimes k[t]$ is the characteristic polynomial of $\phi$.
Proof: Consider how $\text{det}(\Phi)$ factors as $\Phi \circ \text{adj}(\Phi) = \text{adj}(\Phi) \circ \Phi$ in $\text{End}(V)$ for a finite dimensional vector space $V$ with endomorphism $\Phi : V \rightarrow V$. We want a factorization of $p(t)$ of $\phi$ into some polynomial analogous to the adjugate and a linear term $t  \phi$: $$p(t) = f(t)(t  \phi)$$ These two factorizations are analogous, and in fact, if we get the formality right, we can view these as corresponding factorizations in isomorphic rings $\text{End}(V \otimes k[t]) \cong \text{End}(V) \otimes k[t]$.
Construct a map
$$\text{End}_k (V) \otimes_k k[t] \rightarrow \text{Hom}_k(V, V \otimes_k k[t])$$
sending $\phi \otimes t^n$ to the map sending $v$ to $\phi(v)t^n$. This is injective, and surjective since $V$ is finitely generated. Composing these isomorphisms gives an isomorphism $F : \text{End}_{k} (V)[t] \rightarrow \text{End}_{k[t]} (V \otimes_k k[t])$.
View $t  \phi$ as a $k[t]$linear endomorphism of $V \otimes_k k[t]$. Under the isomorphism $F$, $\text{char}(\phi)$ maps to $\text{det} (t  \phi) 1_{V \otimes_k k[t]} )$ and $F ( t  \phi ) = t  \phi$. $t  \phi$ divides $\text{det}(t  \phi) 1_{V \otimes_k k[t]}$ in $\text{End}_{k[t]} (V \otimes_k k[t])$, since $\text{det} (t  \phi) 1_{V \otimes_k k[t]} = \text{adj}(t  \phi) (t  \phi)$, where $\text{adj}(t  \phi)$ is the adjugate matrix. Therefore, $t  \phi$ divides $\text{char}(\phi)$ in $\text{End}_{k}(V)[t]$. So $\text{char}(\phi)$ has $\phi$ as a root in $\text{End}_k(V)$, so that the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$.
The characteristic polynomial $p(t)$ of $\phi \in \text{End}_k (V)$ naturally lives in $\text{End}_k(V)[t]$ from the natural map $\text{End}_k(V) \rightarrow \text{End}_k(V) \otimes_k k[t]$. View $t \text{Id}_V  \phi$ as having endomorphisms as coefficients, and then take the determinant, which is then in $\text{End}_k (V)[t]$.
In the isomorphism $$\text{End}_k ( V \otimes_k k[t]) \cong \text{End}_k (V)[t]$$ We have corresponding elements $$\Phi \leftrightarrow t  \phi$$ and $$\text{det}(\Phi) \leftrightarrow p(t)$$ Therefore, the factorization $\text{det}(\Phi) 1_{V \otimes_k k[t]} = \text{adj}(\Phi) \Phi$ corresponds to a factorization $p(t) = f(t)(t\phi)$ in $\text{End}_k (V) [t]$. And that's the whole idea!

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