[proofplan]
Existence follows from $\dim \mathrm{End}(V) = n^2$ being finite, which forces a linear dependence among powers of $\alpha$ and yields a non-trivial annihilating polynomial. The set of annihilating polynomials forms an ideal of $\mathbb{F}[t]$; uniqueness of the monic generator of minimal degree follows from the leading-term cancellation argument. The divisibility property follows from the division algorithm in $\mathbb{F}[t]$.
[/proofplan]
[step:Construct a non-trivial annihilating polynomial via finite-dimensionality]
Set $n = \dim V$, so $\dim \mathrm{End}(V) = n^2$.
The $n^2 + 1$ endomorphisms $\mathrm{id}, \alpha, \alpha^2, \dots, \alpha^{n^2}$ lie in the $n^2$-dimensional space $\mathrm{End}(V)$, so they are linearly dependent.
There exist $a_0, \dots, a_{n^2} \in \mathbb{F}$, not all zero, with
\begin{align*}
a_0\,\mathrm{id} + a_1 \alpha + \cdots + a_{n^2} \alpha^{n^2} = 0.
\end{align*}
The polynomial $p(t) = a_0 + a_1 t + \cdots + a_{n^2} t^{n^2} \in \mathbb{F}[t]$ satisfies $p(\alpha) = 0$.
Hence the set $\mathcal{I} = \{f \in \mathbb{F}[t] : f(\alpha) = 0\}$ is non-empty.
[guided]
The argument exploits the fact that $\mathrm{End}(V)$ is a finite-dimensional algebra. The evaluation map $\mathrm{ev}_\alpha: \mathbb{F}[t] \to \mathrm{End}(V)$ defined by $t \mapsto \alpha$ is a ring homomorphism sending polynomials to endomorphisms. Since $\mathbb{F}[t]$ is infinite-dimensional but $\mathrm{End}(V)$ has dimension $n^2$, the map $\mathrm{ev}_\alpha$ cannot be injective, so its kernel $\mathcal{I} = \ker(\mathrm{ev}_\alpha)$ is a non-zero ideal of $\mathbb{F}[t]$. To find an explicit element of $\mathcal{I}$, note that the $n^2 + 1$ elements $\mathrm{id}, \alpha, \dots, \alpha^{n^2}$ are linearly dependent in $\mathrm{End}(V)$, yielding a non-trivial polynomial $p$ with $p(\alpha) = 0$.
[/guided]
[/step]
[step:Verify that $\mathcal{I}$ is an ideal of $\mathbb{F}[t]$]
The set $\mathcal{I}$ is non-empty by the previous step.
For $f, g \in \mathcal{I}$: $(f + g)(\alpha) = f(\alpha) + g(\alpha) = 0 + 0 = 0$, so $f + g \in \mathcal{I}$.
For $h \in \mathbb{F}[t]$ and $f \in \mathcal{I}$: $(hf)(\alpha) = h(\alpha) \circ f(\alpha) = h(\alpha) \circ 0 = 0$, so $hf \in \mathcal{I}$.
Here we used that $\mathrm{ev}_\alpha: \mathbb{F}[t] \to \mathrm{End}(V)$ is a ring homomorphism, so evaluation respects products.
Therefore $\mathcal{I}$ is an ideal of $\mathbb{F}[t]$.
[/step]
[step:Prove uniqueness of the monic polynomial of minimal degree]
Among all non-zero polynomials in $\mathcal{I}$, choose one of minimal degree $M(t)$.
Dividing by its leading coefficient, we may assume $M$ is monic.
If $M'$ is another monic polynomial of the same minimal degree with $M'(\alpha) = 0$, then $(M - M')(\alpha) = 0$ and $\deg(M - M') < \deg M$ since the leading terms cancel.
By minimality of $\deg M$, $M - M' = 0$, so $M' = M$.
[/step]
[step:Establish the divisibility property $p(\alpha) = 0 \Leftrightarrow M_\alpha \mid p$]
($\Rightarrow$): Let $p \in \mathbb{F}[t]$ satisfy $p(\alpha) = 0$.
By the division algorithm in $\mathbb{F}[t]$, write $p(t) = q(t)\,M_\alpha(t) + r(t)$ with $\deg r < \deg M_\alpha$ or $r = 0$.
Evaluating at $\alpha$:
\begin{align*}
0 = p(\alpha) = q(\alpha) \circ M_\alpha(\alpha) + r(\alpha) = 0 + r(\alpha) = r(\alpha).
\end{align*}
So $r \in \mathcal{I}$ with $\deg r < \deg M_\alpha$.
By minimality of $\deg M_\alpha$, we must have $r = 0$, giving $M_\alpha(t) \mid p(t)$.
($\Leftarrow$): If $M_\alpha(t) \mid p(t)$, write $p(t) = q(t)\,M_\alpha(t)$.
Then $p(\alpha) = q(\alpha) \circ M_\alpha(\alpha) = q(\alpha) \circ 0 = 0$.
[guided]
The divisibility property says that $\mathcal{I} = (M_\alpha)$, i.e., the ideal of annihilating polynomials is principal, generated by $M_\alpha$. This is a consequence of $\mathbb{F}[t]$ being a principal ideal domain: every ideal is generated by a single element, and the minimal polynomial $M_\alpha$ is precisely this generator, chosen to be monic.
The division algorithm is the key tool. Given $p(\alpha) = 0$, we write $p = qM_\alpha + r$ with $\deg r < \deg M_\alpha$. Evaluating at $\alpha$ forces $r(\alpha) = 0$, but $r$ has degree strictly less than $M_\alpha$, and $M_\alpha$ is the non-zero polynomial of least degree in $\mathcal{I}$. The only polynomial in $\mathcal{I}$ of degree less than $\deg M_\alpha$ is $r = 0$.
[/guided]
[/step]
