if $A^2 = A$ then $|A|=0$ or $|A| =1$ More a verification of work then anything else, I am trying to prove the above statement.  Intuitively I feel that if $A^2 = A$ then $\det(A) = \det(A^2)$. From here I know the property of $\det(AB) = \det(A)\det(B)$ can be used to prove the statement, but I am looking for either a different way to solve it or a confirmation of my intuitive jump from $A^2 = A$ to $\det(A) = \det(A^2)$.
 A: As mentioned above in the comments, by the very definition of a function (equivalently map and operator), if $x=y$ then $f(x)=f(y)$.  I.e. if we plug in the same thing, we expect to get the same result each time.  Now, note that $\det$ is an operator.
As requested though, there is another approach to proving the statement.  I approach using eigenvalues.
As $A^2=A$ we have $A^2-A=0$.  Noting that $A^2-A=A(A-I)=0$ we have that either $\det(A)=0$ or $\det(A-I)=0$.  If it was the first case, we are done.
Suppose then that $\det(A-I)=0$ and $\det(A)\neq 0$.  Then $1$ is an eigenvalue of the matrix and $0$ is not an eigenvalue of the matrix.  There exists then an eigenvector $x$ such that $Ax=1x$.
Suppose there was another eigenvalue, $\lambda$ and associated eigenvector $v$ such that $Av = \lambda v$.  Then we should have $A(Av)=A(\lambda v) = \lambda Av = \lambda^2 v$, but since $A(Av)=A^2v = Av = \lambda v$, we have that $\lambda^2 v = \lambda v$, implying that $\lambda^2 = \lambda$.  Since $0$ is not an eigenvalue, this implies that $\lambda = 1$.  Thus the only eigenvalue is $1$.
Since $\det(A) = \prod \lambda_i$, we have then that $\det(A)=1$

The easier proof is as you seem to suggest: $A^2=A\Rightarrow \det(A^2)=\det(A)=\det(A)\det(A)\Rightarrow \det(A)(\det(A)-1)=0\Rightarrow \det(A)=0$ or $\det(A)=1$
A: if $A^2 = A$ ,
then $|A^2| =|A|$ 
by the property of determinants, $det(AB) = det(A)det(B)$. Therefore
$|A^2| =|A|^2$
so we got
$|A| =|A|^2$
$|A|(1-|A|) =0$
$|A|=0$ or $|A| =1$
