Eigenvalues of $A$ given $I,0\neq A=A^2$ 
Let $A\in \mathbb{M}_{n\times n}^\mathbb{F}$. Prove that if $A\neq 0,I$ and $A^2=A$, then $0,1$ are the only eigenvalues of $A$.

I couldn't really make much progress. I know $A(A-I)=0$, and those two matrices aren't zero. I know $|A|=0$ or  $|A-I|=0$, which is no help at all. Not to mention I need to show that every eigenvalue of $A$ is 0 or 1.
Thanks for any help attemp in advance.
 A: Since $0=A^2-A=((X-1)X)[X:=A]$, all eigenvalues must be roots of the polynomial $(X-1)X$ annihilating $A$, i.e., they must either be $1$ or $0$. And since that polynomial is split with simple roots, $A$ is necessarily diagonalisable. In particular if either $0$ or $1$ is missing as eigenvalue, then one would have a diagonalisable matrix with no other eigenvalues than $1$, whence $A=I$, respectively a diagonalisable matrix with no other eigenvalues than $0$, whence$A=0$, which possibilities were both explicitly forbidden. So both eigenvalues $0,1$ actually occur, and no other eigenvalues do.
A: Let $\lambda$ be an arbitrary eigenvalue of $A$.  There exists an eigenvector $x$ such that $\lambda x = Ax$.  Then we have
\begin{align*}
  \lambda x &= A x\\
    &= A^2 x\\
    &= A(Ax)\\
    &= A(\lambda x)\\
    &= \lambda (A x)\\
    &= \lambda(\lambda x)
\end{align*}
So $\lambda x = \lambda^2x$, giving us $(\lambda - \lambda^2)x = 0$.  Since $x$ is an eigenvector we necessarily have $\lambda - \lambda^2 = 0$.  And $\lambda$ was arbitrary, so...
