If $A$ is a diagonalizable $n\times n$ matrix for which the eigenvalues are $0$ and $1$, then $A^2=A$. 
If $A$ is a diagonalizable $n\times n$ matrix for which the eigenvalues are $0$ and $1$, then $A^2=A$.

I know how to prove this in the opposite direction, however I can't seem to find a way prove this. Could anyone please help?

Write $A = QDQ^{-1}$, where $D$ is a diagonal matrix with the eigenvalues, $0$s and $1$s, on the diagonal. The $A^2 = QDQ^{-1}QDQ^{-1} = QD^{2}Q^{-1}$. But $D^2 = D$, because when you square a diagonal matrix you square the entries on the diagonal and $1^2 = 1$ and $0^2 = 0$. Thus 
$$A^{2} = QD^{2}Q^{-1} = QDQ^{-1} = A$$
