Show that if $M $ is orthogonal matrix, then $\det M =\pm 1$.

Question

I write like following my prove, and I need to know if it's correct:

If P is an orthogonal matrix, then $$PP^t = I = P^tP$$ Therefore, $$\det(P^tP) = \det (I) \implies \det(P) \det (P^t) \implies \det (P) \det (P) = [\det (P)]^2$$ So $$ [\det(P)]^2 = 1 \implies \det (P) = \pm \sqrt 1 $$ So $$ \det (P) = \pm 1$$

I prove it writing this, I would like to know if this proof is correct.

Answer 1

Since $M$ is orthogonal, we have $$MM^T = I$$ It gives $$\det(MM^T) = \det(I) = 1$$ Therefore, $$\det(M) \det(M^T) = 1$$ now since $\det(M^T) = \det(M)$, we get $$\det(M)^2 = 1 \implies \det(M) = \pm 1$$