Prove that every square matrix $A^2 = I$ has at least one of the eigenvalues $\pm 1$ and no other eigen values 
Prove that every matrix $A \in \mathbb{R}^{n \times n}$ with $A^2 = I$
  has at least one of the eigenvalues $\pm 1$ and no other eigenvalues
  except for $\pm 1$.

I'm not sure how to do this correctly?
So we have that $A^2 = I$ where $I$ is the identity matrix and we can form this to
$$A^2 -I = 0$$
Now if we take $x^2 -1$ with $x \in \mathbb{R}$, we have that it's a polynom with simple roots $-1$ and $1$ and no others.
Is it fine like that? :s
 A: Well, if
$A^2 = I, \tag 1$
we may write
$(A + I)(A - I) = A^2 - I = 0; \tag 2$
therefore, for any vector $v \in \Bbb R^n$,
$(A + I)(A - I) v = 0; \tag 3$
now if $(A - I)v = 0$ for all $0 \ne v \in \Bbb R^n$, then
$Av = Iv = 1 \cdot v, \tag 4$
and every $v \ne 0$ is an eigenvector with eigenvalue $1$; if there is some $v$ for which 
$w = (A - I)v \ne 0, \tag 5$
then
$(A + I)w = (A + I)(A - I)v = 0, \tag 6$
or
$Aw = -w, \tag 7$
so $w$ is an eigenvector with eigenvalue $-1$.
So at least one of $\pm 1$ must be an eigenvalue.  To see that there are
no other possibilities, note that if
$Au = \lambda u, \tag 8$
then
$u = Iu = A^2 u = A(Au) = A(\lambda u) = \lambda Au = \lambda^2 u, \tag 9$
so 
$(\lambda^2 - 1) u = 0, \tag{10}$
and since eigenvectors by definition cannot be zero, we have
$\lambda^2 - 1 = 0, \tag{11}$
so
$\lambda = \pm 1 \tag{12}$
are the only possibilities.
Note Added in Edit, Saturday 26 May 2018 10:55 AM PST:  I think it is worth pointing out that this answer generalizes nicely to the case when $A$ is a matrix over a more general field $\Bbb F$, $A \in \Bbb F^{n \times n}$; indeed, the proof is more-or-less the same when $\text{char}(\Bbb F) \ne 2$; when $\text{char}(\Bbb F) = 2$ there is an interesting twist or two which I will leave to my readers to discover.  End of Note.
A: Let A be an $n \times n$ matrix such that $A^2=I$. Let $v$ ben an eigenvector of $A$ by eigenvalue $\lambda$.  Then, $$A^2v=I v=v$$ and $$A^2v=\lambda^2v$$. It follows that $\lambda^2=1$ so $A$ has only eigenvalue 1 and -1.
A: $$\mathbf{0}v = (A^2-I)v = (A-I)\circ(A+I)v$$
Either $(A+I)v = 0$ or if $(A+I)v = z \neq 0$, then $(A-I)z = 0$. Either implies that $-1$ or $1$ is an eigenvalue of $A$.
A: The polynomial $x^2 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $x^2 - 1$.
Therefore
$$\sigma(A) = \{\text{ roots of } m_A\,\} \subseteq \{\text{ roots of } x^2-1\,\} = \{-1,1\}$$
Furthermore, the minimal polynomial must be one of these:
$$x-1, \quad x+1, \quad x^2-1$$
Since the minimal polynomial divides the characteristic polynomial $p_A$, we get that at least one of the factors $x-1$ and $x+1$ divides $p_A$.
Therefore, at least one of $p_A(1)$ and $p_A(-1)$ iz zero, i.e. $1 \in \sigma(A)$ or $-1 \in \sigma(A)$.
