Show that the $\text{Tr}(A)^2 = \text{Tr}(A^2)+\text{Sum of Eigenvalues} $

This is perfectly correct. $$\\$$ As a plus, you can show the "it can be shown that", in a one-line proof: If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, $$ A^2 v = A(Av) = A(\lambda v) = \lambda (Av)= \lambda^2 v $$