1. **Reduction to the case $k = 2$**:
- We start by noting that it suffices to prove the result for $k = 2$. If $\det A = b_1 b_2 \cdots b_k$, we can write $\det A = (b_1 \cdots b_{k-1}) b_k$. Then, we can find integer matrices $B_{k-1}'$ and $B_k$ such that $A = B_{k-1}' B_k$ with $\det B_{k-1}' = b_1 \cdots b_{k-1}$ and $\det B_k = b_k$. We can repeat this process iteratively.

2. **Case $k = 2$**:
- Suppose $A$ is an $n \times n$ matrix with integer entries and $\det A = bc$ for some integers $b$ and $c$. We need to show that there exist $n \times n$ integer matrices $B$ and $C$ such that $A = BC$ and $\det B = b$, $\det C = c$.

3. Upper Triangular Matrices:
- Multiplying a matrix $A$ by elementary matrices representing row-addition operations does not alter the determinant of $A$. If $A' = E A$ is a row-reduced form of $A$ with $E$ a product of elementary matrices, then $\det E = 1$ and $E$ has integer entries. Hence, it suffices to prove the problem for upper triangular matrices.

4. **Induction on $n$**:
- Suppose $A$ is an upper triangular matrix with integer entries and $\det A = bc$. We will use induction on $n$.

5. **Base Case $n = 2$**:
- Consider $A = \begin{pmatrix} m & x \\ 0 & n \end{pmatrix}$. We need to find $B$ and $C$ of the form:
$$B = \begin{pmatrix} t_1 & y \\ 0 & t_2 \end{pmatrix}, \quad C = \begin{pmatrix} \frac{m}{t_1} & z \\ 0 & \frac{n}{t_2} \end{pmatrix}$$
for some integers $t_1, t_2, y, z$ such that $t_1 \mid m$ and $t_2 \mid n$. We have $\det A = mn = bc$ and thus we need $t_1 t_2 = b$. Finally, $y$ and $z$ must satisfy $x = t_1 z + \frac{n}{t_2} y$. For a given $x$, we can guarantee a solution $(y, z)$ exists only if $\gcd(t_1, \frac{n}{t_2}) = 1$.

6. Lemma:
- If $b \mid mn$, then we can find $t_1 \mid m$ and $t_2 \mid n$ such that $b = t_1 t_2$ and $\gcd(t_1, \frac{n}{t_2}) = 1$.
- Proof of Lemma: Assume $mn$ and $b$ have the same set of prime factors. It suffices to prove the lemma when $b$, $m$, and $n$ are all powers of the same prime. If $b \mid n$, choose $t_1 = \frac{b}{n}$ and $t_2 = n$. If $n \mid b$, choose $t_1 = 1$ and $t_2 = b$.

7. Inductive Step:
- Suppose $A$ is an $n \times n$ matrix and let $A'$ be the lower-right $(n-1) \times (n-1)$ submatrix of $A$. Denote $m = A_{1,1}$ and $n = \det A'$. In block form:
$$A = \begin{pmatrix} m & X \\ 0 & A' \end{pmatrix}$$
Since $A$ is upper triangular, $\det A = A_{1,1} \det A' = mn = bc$. By the lemma, we can find $b_1, c_1, b_2, c_2$ such that $m = b_2 c_2$ and $n = b_1 c_1$ with $\gcd(b_2, c_1) = 1$. By the inductive hypothesis, we can find $(n-1) \times (n-1)$ matrices $B'$ and $C'$ such that $A' = B' C'$ and $\det B' = b_1$, $\det C' = c_1$.

8. **Constructing $B$ and $C$**:
- Take $B$ and $C$ in block form:
$$B = \begin{pmatrix} b_2 & Y \\ 0 & B' \end{pmatrix}, \quad C = \begin{pmatrix} c_2 & Z \\ 0 & C' \end{pmatrix}$$
We need row vectors $Y$ and $Z$ such that $b_2 Z + Y C' = X$. Write $Y = [y_1, \ldots, y_{n-1}]$ and $Z = [z_1, \ldots, z_{n-1}]$ and $X = [x_1, \ldots, x_{n-1}]$.

9. **Solving for $Y$ and $Z$**:
- Matching the first coordinate, we have $b_2 z_1 + y_1 C'_{1,1} = x_1$. There exists an integer solution $(y_1, z_1)$ because $\gcd(b_2, C'_{1,1}) \mid \gcd(b_2, \det(C')) = \gcd(b_2, c_1) = 1$.
- Matching the second coordinate, we have $b_2 z_2 + y_1 C'_{1,2} + y_2 C'_{2,2} = x_2$, which becomes $b_2 z_2 + y_2 C'_{2,2} = x_2 - y_1 C'_{1,2}$. This admits a solution $(y_2, z_2)$ for the same reason as above.
- Continue in this manner to determine $Y$ and $Z$ satisfying the relation. The key is that $b_2$ is relatively prime to $c_1 = \det C'$ and thus to all the diagonal entries of $C'$.

$\blacksquare$