Answer
See the proof below.
Work Step by Step
Consider the following statement:
For every positive integer $n$, given square matrices $A_1,A_2,A_3,\ldots,A_n$ of the same size we have that
$$\left|A_1A_2A_3\cdots A_n\right|=\left|A_1\right|\left|A_2\right|\left|A_3\right|\cdots\left|A_n\right|.$$
We will use mathematical induction to prove that the statement is true.
$\textbf{1.}$ We have that the statement is true for $n=1$ since $\left|A_1\right|=\left|A_1\right|$.
$\textbf{2.}$ Suppose the statement holds for $n=k$, that is, suppose that
$$\left|A_1A_2A_3\cdots A_k\right|=\left|A_1\right|\left|A_2\right|\left|A_3\right|\cdots\left|A_k\right|.$$
where the matrices $A_1,A_2,\ldots,A_k$ are square matrices of the same size.
Now, let $A_1,A_2,A_3,\ldots,A_k,A_{k+1}$ be square matrices of the same size.
We have that
$$\left|A_1A_2A_3\cdots A_kA_{k+1}\right|=\left|\left(A_1A_2A_3\cdots A_k\right)A_{k+1}\right|.$$
Now, we know from a previous theorem that if $A$ and $B$ are square matrices of the same size then, it is true that $\left|AB\right|=\left|A\right|\left|B\right|$ and therefore, if we let
$A=A_1A_2A_3\cdots A_k$ and $B=A_{k+1}$ we have that
$$\left|\left(A_1A_2A_3\cdots A_k\right)A_{k+1}\right|=\left|A_1A_2A_3\cdots A_k\right|\left|A_{k+1}\right|.$$
By our hypothesis we have that
$$\left|A_1A_2A_3\cdots A_k\right|=\left|A_1\right|\left|A_2\right|\left|A_3\right|\cdots\left|A_k\right|$$
which implies that
$$\left|A_1A_2A_3\cdots A_k\right|\left|A_{k+1}\right|=\left|A_1\right|\left|A_2\right|\left|A_3\right|\cdots\left|A_k\right|\left|A_{k+1}\right|$$
and therefore,
$$\left|A_1A_2A_3\cdots A_kA_{k+1}\right|=\left|A_1\right|\left|A_2\right|\left|A_3\right|\cdots\left|A_k\right|\left|A_{k+1}\right|.$$
Thus, statement is true for $n=k+1$ whenever it is true for $n=k$.
Hence, the mathematical induction principle guarantees that the statement is true for every positive integer $n$.
