A better proof for $\det(P) = \pm1$ if $P$ is an orthogonal matrix

Here is a geometric argument that can be used for real orthogonal matrices: If $T:\ V\to V$ is an arbitrary linear transformation of a finite dimensional euclidean vector space $V$ then the volumes of arbitrary measurable sets $A\subset V$, in particular of balls or cubes, are multiplied by $\bigl|\det(T)\bigr|$. That is to say, one has $${\rm vol}\bigl(T(A)\bigr)=\bigl|\det(T)\bigr|\ {\rm vol}(A)\ .$$ Now an orthogonal transformation $T$ transforms the unit ball $B:=\bigl\{x\in{\mathbb R}^n\bigm| |x|\leq1\bigr\}$ onto itself, and ${\rm vol}(B)>0$. It follows that $|\det(T)\bigr|=1$.

The absolute value of the determinant of a real, square matrix is the volume of the parallelepiped whose sides are the columns of the matrix. For an orthogonal matrix, the columns are pairwise orthogonal unit vectors, so this parallelepiped is a cube with side length 1. So the volume is 1, and therefore the determinant is $\pm1$.