Determinant of Matrix Product/General Case

Theorem
Let $\mathbf {A_1}, \mathbf{A_2}, \cdots, \mathbf{A_n}$ be square matrices of the same size, where $n > 1$.

Then:


 * $\det \left({\mathbf {A_1} \mathbf {A_2} \cdots \mathbf {A_n}}\right) = \det \left({\mathbf {A_1}}\right) \det \left({\mathbf {A_2}}\right) \cdots \det \left({\mathbf {A_n}}\right)$

where $\det \mathbf A$ denotes the determinant of $\mathbf A$.

Proof
Proof by induction:

Base case
$n = 2$ holds by Determinant of Matrix Product.

So shown for base case.

Induction Hypothesis
This is our induction hypothesis:
 * $\det \left({\mathbf {A_1} \mathbf {A_2} \cdots \mathbf {A_k}}\right) = \det \left({\mathbf {A_1}}\right) \det \left({\mathbf {A_2}}\right) \cdots \det \left({\mathbf {A_k}}\right)$

Now we need to show true for $n=k+1$:
 * $\det \left({\mathbf {A_1} \mathbf {A_2} \cdots \mathbf {A_k} \mathbf {A_{k+1}}}\right) = \det \left({\mathbf {A_1}}\right) \det \left({\mathbf {A_2}}\right) \cdots \det \left({\mathbf {A_k}}\right) \det \left({\mathbf {A_{k+1}}}\right)$

Induction Step
This is our induction step:

The result follows by induction.

Therefore:
 * $\det \left({\mathbf {A_1} \mathbf {A_2} \cdots \mathbf {A_n}}\right) = \det \left({\mathbf {A_1}}\right) \det \left({\mathbf {A_2}}\right) \cdots \det \left({\mathbf {A_n}}\right)$