Answer
See the proof below.
Work Step by Step
Consider the following statement:
For every positive integer $n$, given matrices $A_1,A_2,\ldots,A_n$ with sizes such that the product $A_1A_2A_3\cdots A_n$ is defined, we have that
$$\left(A_1A_2A_3\cdots A_n\right)^T=A_n^T\cdots A_3^TA_2^TA_1^T.$$
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 $A_1^T=A_1^T$.
$\textbf{2.}$ Suppose the statement holds for $n=k$, that is, suppose that
$$\left(A_1A_2A_3\cdots A_k\right)^T=A_k^T\cdots A_3^TA_2^TA_1^T$$
where the matrices $A_1,A_2,\ldots,A_n$ are such that the product $A_1A_2A_3\cdots A_n$ is defined.
Now, let $A_1,A_2,A_3,\ldots,A_k,A_{k+1}$ be matrices with sizes such that the product $A_1A_2A_3\cdots A_kA_{k+1}$ is defined.
Then, we have that
$$\left(A_1A_2A_3\cdots A_kA_{k+1}\right)^T=\left(\left(A_1A_2A_3\cdots A_k\right)A_{k+1}\right)^T.$$
Now, we know from a previous theorem that if $A$ and $B$ are matrices such that the product $AB$ is defined, then $\left(AB\right)^T=B^TA^T$ and therefore, if we let
$A=A_1A_2A_3\cdots A_k$ and $B=A_{k+1}$ we have that
$$\left(\left(A_1A_2A_3\cdots A_k\right)A_{k+1}\right)^T=A_{k+1}^T\left(A_1A_2A_3\cdots A_k\right)^T.$$
By our hypothesis we have that
$$\left(A_1A_2\cdots A_k\right)^T=A_k^T\cdots A_3^TA_2^TA_1^T$$
which implies that
$$A_{k+1}^T\left(A_1A_2A_3\cdots A_k\right)^T=A_{k+1}^T\left(A_k^T\cdots A_3^TA_2^TA_1^T\right)=A_{k+1}^TA_k^T\cdots A_3^TA_2^TA_1^T$$
and therefore,
$$\left(A_1A_2A_3\cdots A_kA_{k+1}\right)^T=A_{k+1}^TA_k^T\cdots A_3^TA_2^TA_1^T.$$
Thus, the statement is true for $n=k+1$ whenever it is true for $n=k$.
Hence, the mathematical induction principle guarantees that the formula is true for every positive integer $n$.
