For an orthogonal matrix $Q$, why does $QQ^T = I$?

From $\mathrm Q^{\top} \mathrm Q = \mathrm I_n$, we conclude that $\mathrm Q^{\top}$ is the left inverse of $\mathrm Q$. Since $\mathrm Q$ is square, the left inverse is also the right inverse and, thus, $\mathrm Q \mathrm Q^{\top} = \mathrm I_n$. Hence, the inverse of $\mathrm Q$ is $\mathrm Q^{-1} = \mathrm Q^{\top}$.

If we know that $Q^TQ=I$ we may note that $$ QQ^T = (QQ^T)(QQ^{-1}) = Q(Q^TQ)Q^{-1} = QIQ^{-1} = I. $$ Here we used the fact that matrix $Q^{-1}$ exist which is follows from $Q^TQ = I$ (indeed, $\det (QQ^T) = 1 \Rightarrow (\det Q)^2 = 1\Rightarrow \det Q \neq 0)$ .