Answer
See below
Work Step by Step
Let $A$ be an $n \times n$ orthogonal matrix.
Obtain $A^T=A^{-1}$ which gives us that $AA^T=I_n$ and then $\det(AA^T)=\det(I_n)$.
According to P8, we have $\det(AA^T)=\det(A)\det(A^T)$
and by P4, $\det(A^T)=\det(A)$.
As the result, $\det(AA^T)=\det(A)\det(A)=(\det(A))^2$
Since $\det(I_n)=1 \rightarrow (\det(A))^2=1\rightarrow \det (A)=\pm 1$
