= Solution
The characteristic <polynomial> is
$$
\chi_A(t)=\det(tI-A).
$$
To prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb C$, $\chi_A$ has a root $\lambda$, so $A$ has an <eigenvector> $v_1$. Extend it to a <basis>. In this <basis>,
$$
[A]=\begin{pmatrix}\lambda&*\\0&B\end{pmatrix}.
$$
By induction, a change among the remaining $n-1$ <basis> <vectors> makes $B$ upper triangular. Thus the <triangularization over an algebraically closed field> gives
$$
\boxed{A\text{ is similar to an upper-triangular matrix}}.
$$
The minimal <polynomial> $m_A$ is the monic <polynomial> of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular <matrix> similar to $A$, with diagonal entries $\lambda_1,\ldots,\lambda_n$, and put
$$
V_k=\operatorname{span}\{e_1,\ldots,e_k\}.
$$
Then
$$
(T-\lambda_kI)V_k\subseteq V_{k-1}.
$$
The factors $T-\lambda_kI$ commute, so applying all of them successively lowers the invariant flag to zero:
$$
\prod_{k=1}^n(T-\lambda_kI)=0.
$$
Similarity gives the same <polynomial> identity for $A$. Hence a nonzero monic annihilating <polynomial> of degree $n$ exists, and a least-degree one exists.
If $m$ and $\widetilde m$ were two monic annihilating <polynomials> of the same least degree, then $m-\widetilde m$ would be an annihilating <polynomial> of smaller degree unless it were zero. Thus the minimal <polynomial> is unique, and the <minimal polynomial bound from a triangular invariant flag> gives
$$
\boxed{\deg m_A\leq n}.
$$
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