Since $A^2=A$, one can create the isomorphism
$$V \cong \text{Im } A \oplus \ker A$$
$$ x \mapsto (Ax,(I-A)x).$$

To show that this is an isomorphism is simple. Hence, we have your item $2$.

The decomposition above also shows that the unique eigenvalues are $0$ or $1$. The multiplicity of the eigenvalue $1$ will give the rank of the operator. Since the trace is the sum of the eigenvalues (counted with multiplicity), you have your item $1$.

EDIT: Since this answer was downvoted but not explained, I will assume it is due to the need of clarification with respect to the fact that I am conflating the algebraic multiplicity and the geometric multiplicity:

Both coincide, since $A$ is diagonalizable: it is clear that $A$ restricts to the identity on its image on the decomposition above, and thus any basis $\{v_1,\cdots,v_j\}$ of $\mathrm{Im}~A$ is composed of eigenvectors. It is obvious that any basis $\{v_{j+1},\cdots,v_n\}$ of $\ker A$ is also of eigenvectors, thus $\{v_1,\cdots,v_n\}$ is a basis of $V$ composed of eigenvectors for $A$.

If anything else needs clarification, please feel free to point it out.