Prove that the characteristic polynomial of $$A = \begin{bmatrix} B & 0 \\ 0 & C \end{bmatrix}$$ where $B$ and $C$ are square matrices, is the product of the characteristic polynomials of $B$ and $C$.
My attempt is the following. Let $A \in M_{n\times n}(\mathbb{F})$ the polynomial $f(t)=\det(A-\lambda I_n)$ is the characteristic polynomial for $A$. I´m not sure if what should be done is to prove that the determinant of a matrix $A$ is equal to the product of its eigenvalues?
$$ A=\begin{bmatrix} B & 0 \\ 0 & C \end{bmatrix}, $$
$$ P_A(\lambda)=(A-\lambda I), $$ $$ P_A(\lambda)=\det(\begin{bmatrix}B-\lambda & 0 \\ 0 & C - \lambda \end{bmatrix}), $$ $$ P_A(\lambda)= (B-\lambda)(C-\lambda)-0, $$ $$ =BC-B\lambda -C\lambda +\lambda^2 $$ $$ P_A(\lambda)=\lambda^2 -\lambda(B+C) + BC. $$
Now, if I have $B \in M_{n\times n}(\mathbb{F})$ the polynomial $f(t)=\det(B-\lambda I_n)$ is the characteristic polynomial for $B$ and the same for $C$, $C \in M_{n\times n}(\mathbb{F})$ where its characteristic polynomial is $f(t)=\det(C-\lambda I_n)$.
The product of its characteristic polynomials is:
$$ (B-\lambda I_n)(C-\lambda I_n) $$ $$ (B-\lambda)(C-\lambda) $$ $$ =BC-B\lambda -C\lambda +\lambda^2 $$
But what do I do with the identity?
Please let me know if what I did is correct for the problem.