Eigenvalue of a matrix Let $A$ be an $n\times n$ matrix and let $I$ be the $n\times n$ identity matrix. Show that if $A^{2} = I$, and $A \neq I$, then $\lambda =-1$ is an eigenvalue of $A$. This problem doesn't seem that too hard to solve, but I am stuck near the end. Here is what I have done so far.
Since $A^{2}=I$, then by definition $Ax=\lambda x$, where $x$ is an eigenvector of $A$ and $\lambda$ is an eigenvalue of $A$. It follows that $x=Ix=A^{2}x=A(Ax)=A(\lambda x)= \lambda(Ax)=\lambda^{2}x$. (Now I was going to use the fact that since $A \neq I$, that $x\neq 0$, so we get that $\lambda = 1$ or $-1$). This is where I am stuck. 

Hint: If $-1$ is not an eigenvalue of $A$, then $A+I$ is invertible.  What is $(A^2-I)(A+I)^{-1}$?
