Let $M\in \text{SO}(3,\mathbb{R})$, prove that $\det(M-I\_3)=0$.

My attempt:

$$
\begin{align} 
\det(M-I\_3)&=\det(M-M^TM)\\&=\det((I\_3-M^T)M)\\&=\underbrace{\det(M)}\_{=1}\det(I\_3-M^T)
\end{align} 
$$

Hence $$
\begin{align} 
\det(M-I\_3)&=\det(I\_3-M^T)\\&=\det((I\_3-M)^T)\\&=\det(I\_3-M)\\&=\det(-(M-I\_3))\\&=\underbrace{(-1)^3}\_{=-1}\det(M-I\_3),
\end{align} 
$$ and thus $\det(M-I\_3)=0.$

Is this proof correct or did I miss out on something?