I'm looking into the proof of Cayley Hamilton theorem from multiple sources. All the proofs I saw use the fact that
$$ (A - \lambda I ) \,\mbox{adj} (A - \lambda I ) = | A - \lambda I | I $$
The proof is not very hard but a bit involved to my taste. Here is my idea for an alternative proof:
Let $A$ be a square matrix of order $n$. $A$ is similar to its Jordan Normal Form matrix $J$ namely $ A = P J P^{-1} $ where $P$ is invertible. It is easy to show that if $J$ satisfies Cayley Hamilton then the same is true also for $A$ (I'm skipping the proof for this as it is straight forward).
So we are left to prove that $J$ satisfies Cayley Hamilton, namely if $J$ has $k$ distinct eigenvalues and $\lambda_i $ is an eigenvalue of $J$ with multiplicity of $m_i$ then
$$\prod_{i=1}^{k} (J - \lambda_i I)^{m_i} = 0$$
The matrix $J$ consists of Jordan blocks, one or more blocks for each distinct eigenvalue and its eigenvalues are on the main diagonal. Let $B_i$ be the $m_i \times m_i$ Jordan matrix combining all the Jordan blocks corresponding to $\lambda_i$ in $J - \lambda_i I$. The entries of $B_i$ are $0$ except maybe the entries on the superdiagonal which can be either $0$ or $1$( depending on the exact form of $J$). It is not hard to prove that $B_i^{m_i} = 0 $ (I'm skipping this as well).
If we look at the expression $\Pi_{i=1}^{k} (J - \lambda_i I)^{m_i} $, we see that each $ (J - \lambda_i I)^{m_i} $ term nullifies the sub matrix $B_i$ and the whole expression evaluates to zero as we wanted to prove.
This is just a proof idea. Do you think it is valid ? If yes I wonder why is not used more to prov the Cayley Hamilton theorem ?