linear algebra - Characteristic polynomial of a matrix in block form

Let $A_1, ...,A_n$ be square matrices over a field $F$, and let $A$ have the block form
$$A= \begin{bmatrix} A_1 & \\ & A_2 & & \\ & &... & \\ & & & A_n \end{bmatrix}$$
where all other off-diagonal entries are zero. Show that characteristic polynomial $\Delta_A$ of A is $\Delta_A = \Delta_{A_1} \Delta_{A_2} ... \Delta_{A_n}$

Firstly, I observed that $\Delta_A(A) = 0 \quad \Leftrightarrow \Delta_A(A_1)=...=\Delta_A(A_n) =0$. But I did not proceed from here.

Answer

The characteristic polynomial is a determinant.
$$\lambda I - A = \begin{bmatrix} \lambda I_1 - A_1 & \\ & \lambda I_2 -A_2 & & \\ & &... & \\ & & & \lambda I_n -A_n \end{bmatrix}$$

Where $I_j$ are identity matrices of the same size as $A_j$. The determinant of a block-diagonal matrix is the product of the blocks' determinants:
$$k_A(\lambda) = \det (\lambda I - A) = \det(\lambda I_1 - A_1) \det(\lambda I_2 - A_2) \cdot \ldots \cdot \det(\lambda I_n - A_n).$$