Sum of coeficients

You start with the requirement that the wave function is normalized: $$ langle psi |psirangle = 1 $$

There is a subtle difference between $c^2$ and $vert cvert^2$ since $c^*=langlepsivert e_nrangle$. In particular, you can use the resolution the identity $hat{mathbb{I}}=sum_n vert e_nranglelangle e_nvert$.

$|psi rangle = sum_n c_n |e_n rangle$. Now by normalization $1 = langle psi|psi rangle = sum_{m,n} c_n^star c_m langle e_n|e_m rangle$. Can you finish it from here?

Start from $$ |c_n|^2= langle psi|nrangle langle n |psirangle $$ (this is not what you have) Then, by the spectral theorem for self-adjoint opertors, the states $|nrangle$ form a orthonormal complete set so $$ |psirangle= sum_n|nranglelangle n|psirangle, $$ for any state $|psirangle$. This is often written as $$ {rm id}= sum_n |nrangle langle n |, $$ where ${rm id}$ is the identity operator. The dual "bra" state is expanded as $$ langle psi|= sum_nlangle psi|nranglelangle n| $$ Thus $$ sum_n |c_n|^2= sum_{n,m} langle psi|nrangle langle n |mranglelangle m| psirangle =sum_{n} langle psi|nrangle langle n| psirangle = langle psi|psirangle. $$ We have used $langle n|mrangle= delta_{nm}$, so this result does not hold for a general operator $Q$ -- even if it is diagonalizable. The operator has to be self-adjoint (hermitian) so that the eigenstates are mutually orthogonal.