Answer
See below
Work Step by Step
Assume $A$ is a $n\times n$ orthogonal matrix
Then it gives $AA^T=I_n \rightarrow \det(AA^T)=\det(I_n)$.
We know that $\det(AA^T)=\det(A)\det(A^T)$
and $\det(A^T)=\det(A)$
Since $\det(I_n)=1$ then $\det(AA^T)=\det(A)\det(A)=(\det(A))^2=1$
As a result, $\det(A)=\pm 1$
