Determinant of a Block Diagonal Matrix

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Here is Prob. 18, Sec. 2.2, in the book Linear Algebra With Applications by Steven J. Leon and Lisette de Pillis, tenth edition:

Let $A$ be a $k \times k$ matrix and let $B$ be an $(n-k) \times (n-k)$ matrix. Let $$ E = \left[ \begin{matrix} I_k & O \\ O & B \end{matrix} \right], \qquad F = \left[ \begin{matrix} A & O \\ O & I_{n-k} \end{matrix} \right], \qquad C = \left[ \begin{matrix} A & O \\ O & B \end{matrix} \right], $$ where $I_k$ and $I_{n-k}$ are the $k \times k$ and $(n-k) \times (n-k)$ identity matrices.

(a) Show that $\det(E) = \det (B)$. [This is easy. We successively use the cofactor expansions along the first row of each successive non-zero determinant until we reach $\det(B)$. Am I right?]

(b) Show that $\det(F) = \det(A)$. [We successively use the cofactor expansion along the last row of each successive non-zero determinant until we arrive at $\det(A)$. Am I right?]

(c) Show that $\det(C) = \det(A)\det(B)$. [How to do this?]

Here we have $$ E = \left[ \begin{matrix} I_k & O_{k \times (n-k)} \\ O_{(n-k)\times k} & B \end{matrix} \right], \qquad F = \left[ \begin{matrix} A & O_{k \times (n-k)} \\ O_{(n-k)\times k} & I_{n-k} \end{matrix} \right], \qquad C = \left[ \begin{matrix} A & O_{k \times (n-k)} \\ O_{(n-k)\times k} & B \end{matrix} \right], $$ of course, where $O_{k \times (n-k)}$ and $O_{(n-k) \times k}$ denote the zero matrices of order $k \times (n-k)$ and $(n-k) \times k$, respectively.