How to prove the determinant is $|A|^m$?

Question

$A=(a_{ij})$, and $I_m$ is the identity matrix of order $m$. How to prove the following?

$$\left|\begin{array}{cccc}a_{11}I_m&a_{12}I_m&\cdots&a_{1n}I_m\\a_{21}I_m&a_{22}I_m&\cdots&a_{2n}I_m\\\vdots&\vdots&\ddots&\vdots\\a_{n1}I_m&a_{n2}I_m&\cdots&a_{nn}I_m\end{array}\right|=|A|^m$$

Answer 1

Shuffle the rows and columns until the matrix looks like $$ \begin{pmatrix} A & O & \dots & O\\ O & A & \dots & O\\ \vdots & \vdots & \ddots & O\\ O & O & \dots & A\\ \end{pmatrix} $$ and now it is clear.

Answer 2

From the determinant of the Kronecker product,

$$\det \left( \mathrm A \otimes \mathrm I_m \right) = \left( \det \left( \mathrm A \right) \right)^m \cdot \left( \det \left( \mathrm I_m \right) \right)^n = \left( \det \left( \mathrm A \right) \right)^m$$