Linear-algebra first course problem about orthogonal matrices I am trying to demonstrate next assert about matrices:
$A$ is a matrix of $n$ order, with $n$ odd, that obeys $A A^T =I$ and $\det\, A=1$. Then $\det\,(A-I)=0$.
I have tried a number of things but none of them work. That $n$ is odd seems to indicate to the trace of the matrix an its properties but I have also tried to find a product of matrix useful but it did not work out.
All ideas well be apreciated.

We know that $\|Ax\|^2=\langle Ax,Ax\rangle=\langle AA^Tx,x\rangle=\|x\|^2$.  Since $n$ is odd, we know that there’s at least one zero of the characteristic polynomial of $A$, name it $\lambda$ and let $v$ be a correspondent eigenvector.  Then
$$\|v\|=\|Av\|=\|\lambda v\|=|\lambda|\|v\|,$$
hence $\lambda=\pm1$, that is, any real eigenvalue is either $1$ or $-1$.  
Since $\det(A)$ is the product of all eigenvalues of $A$ and that determinant equals $1$, the multiplicity of eigenvalue $-1$ must be even, say $2k$. Then the characteristic polynomial factorized over the real numbers is $(\lambda+1)^{2k}p(\lambda)$ where $p$ has an odd degree and therefore a zero, which must be $1$. That means $\det(A-1\cdot I)=0$.
