Let $A\in M_n(F)$ and $k=\deg(m_A)$ where $m_A$ is the minimal polynomial of $A$.
Prove that $\mathrm{span}\{ I,A,A^2... \} = \mathrm{span} \{ I,A,A^2,..., A^{k-1}\}$
So we have that $m_A = a_0 + a_1x + ... +x^{k}$ (Notice that the coefficient of $x^{k}$ is $1$)
Since, $m_A$ is the minimal polynomial:
$$m_A(A) = 0 \\a_0I+a_1A + ... + A^{k} = 0$$
What should I do next? I am kinda stuck
Since the degree of $m_A$ is $k$ then let $m_A=a_0+a_1x+\cdots+a_{k-1}x^{k-1}\color{red}{+x^k}$ and since it annihilates the matrix $A$ then $$A^k=-a_0 I_n-a_1 A-\cdots-a_{k-1}A^{k-1}\in \operatorname{span} \{ I,A,A^2,..., A^{k-1}\}$$ and by easy induction we get $$A^{k+p}\in \operatorname{span} \{ I,A,A^2,..., A^{k-1}\}\quad\forall p\ge0$$
Since $m_A\in F[x]$ has degree$~k$, the fact that $m_A[A]=0$ establishes that $A^k\in S$ where $S\subseteq M_n(F)$ is the subspace $\operatorname{span}\{A^0,\ldots,A^{k-1}\}$. It immediately follows that $S$ is stable under (say left-) multiplication by$~A$ (check this on the generating vectors of the subspace), so that in particular $A^i\in S$ for all $i\in\Bbb N$.
Note that this only uses that $m_A$ annihilates $A$ (and has degree$~k$), not that it is the minimal polynomial. The latter does imply the stronger statement that $A^0,\ldots,A^{k-1}$ is a basis for $S$.
Hint: Prove by induction that $A^i$ can be written as a linear combination of $I,A,A^2,\dots,A^{k-1}$.
More generally, show that if $B\in\mathrm{span}\{I,A,A^2,\dots,A^{k-1}\}$ then $BA\in\mathrm{span}\{I,A,A^2,\dots,A^{k-1}\}$.
