Determinant of a matrix that is its own inverse I need help in showing that when computing the determinant the inverse   of an $n \times n$ matrix with the property  
$$M=M^{-1}$$
that is
$$M^2 = I$$ 
the determinant is either $1$ or $-1$.
I've tried showing it in a couple ways and the way I'm trying to show it has me stuck
$$K^2 = I$$
$$K^2 - I = 0$$
$$\det(K^2 - I) = 0$$
$$\det(I - I) = 0$$
I get here and I am hopelessly stuck. Could I go on to prove it this way? Is there any  elementary way to prove this?
 A: If
$M = M^{-1}, \tag 1$
then
$M^2 = I, \tag 2$
so by the multiplicative property of determinants,
$(\det M)^2 = \det (M^2) = \det I = 1, \tag 3$
which implies that
$\det M = \pm 1. \tag 4$
Now in fact, we can go a little further with only a little more work and show that every eigenvalue or $M$ is in the set $S = \{-1, 1\}$.  For if
$Mv = \mu v \tag 5$
for some non-zero vector $v$, then
$\mu^2 v = \mu(\mu v) = \mu Mv = M(\mu v) = M(Mv)= M^2 v = Iv = v; \tag 6$
thus
$\mu^2 = 1, \tag 7$
or 
$\mu = \pm 1. \tag 8$
Since the eigenvalues of $M$ lie in the set $S$, and $\det M$ is the product of its eigenvalues, we again see that we must have (4).
Finally, we can also write
$(M + I)(M - I) = M^2 - I = 0, \tag 9$
whence
$\det(M + I) \det(M - I) = 0; \tag {10}$
thus
$\det(M + I) = 0 \tag{11}$
or
$\det(M - I) = 0; \tag{12}$
in the former case, there exists a vector $v$ with
$Mv = -v; \tag{13}$
in the latter
$Mv = v, \tag{14}$
which gives a quick and easy proof of the existence of eigenvectors corresponding the eigenvalues $\mu = \pm 1$.
A: We have that 
$$1 = \det(I) = \det(AA^{-1}) = \det(AA) = \det(A)\det(A) = \det(A)^2.$$ So, $\det(A)^2 - 1 = 0$. This is a polynomial in $\det(A)$, what are its solutions?
