I would suggest prove the Hamilton Cayley theorem in one sentence using algebraic geometry: The theorem is true for diagonalizable matrices, which forms a Zariski dense set. (although it is not open, but it contains the open subset of matrices with distinct eigenvalues. Also, the irreducibility of $\mathbb{A}^n$ is assumed.)
This proof can be compared with the "physicists'" proof of this theorem (over $\mathbb{C}$): If $A$ is a square matrix, then it is diagonalizable after perturbation, i.e. $A+\epsilon H$ is diagonalizable. And let $\epsilon\rightarrow 0$.