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?

Another approach, more theoretical but perhaps simpler and shorter:
Since a matrix is diagonalizable iff its minimal polynomial splits as a product of different linear factors, being that only $\,0,1\,$ are the only eigenvalues of the matrix, its minimal polynomial must be $\,x(x-1)=x^2-x\,$ , from where
$$A^2-A=0\implies A^2=A$$
