Identity in vector calculus : why is it zero?

From Spiegel's "Vector Analysis", problem 7.21 a):

We have general curvilinear coordinates, $$u_i$$, with unit vectors $$\mathbf{e}_i$$ for $$i=1,2,3$$ and a vector $$\mathbf{A}=A_1\mathbf{e}_1+A_2\mathbf{e}_2+A_3\mathbf{e}_3$$ defined in these coordinates. The problem deals with finding the divergence of $$\nabla\cdot(A_1\mathbf{e}_1)$$ We must know that $$h_i$$ is defined as a scaling term to normalize $$\mathbf{e}_i$$, i.e.

$$\frac{\partial\mathbf{r}}{\partial u_i}=h_i\mathbf{e}_i, \hspace{1cm} h_i=\left|\frac{\partial \mathbf{r}}{\partial u_i}\right|$$

We also assume right-handed orthogonal coordinates. Hence, $$\mathbf{e}_1 = \mathbf{e}_2\times \mathbf{e}_3$$ Furthermore, $$u_i=c$$ is also a defined as a surface, and the gradient$$\nabla u_i$$ (is normal to that surface) is parallel to the coordinate $$u_i$$, and can be shown to be $$\nabla u_i = \frac{1}{h_i}\mathbf{e}_i$$

So, if $$\mathbf{e}_1=\mathbf{e}_2\times\mathbf{e}_3$$, then $$\nabla\cdot(A_1\mathbf{e}_1)=\nabla\cdot(A_1h_2h_3\nabla u_2\times\nabla u_3)$$ Using $$\nabla\cdot(\psi\mathbf{A})=\nabla \psi \cdot \mathbf{A}+\psi \nabla\cdot \mathbf{A}$$, we get $$\nabla\cdot(A_1h_2h_3\nabla u_2\times\nabla u_3) =\nabla(A_1h_2h_3)\cdot(\nabla u_2\times \nabla u_3)+(A_1h_2h_3)\nabla\cdot (\nabla u_2\times \nabla u_3)$$ The rightmost term is then said to be zero $$(A_1h_2h_3)\nabla\cdot (\nabla u_2\times \nabla u_3)=0$$ So my question is how can this term always be zero ? We have $$\nabla\cdot\left(\frac{\mathbf{e}_2}{h_2}\times\frac{\mathbf{e}_3}{h_3}\right)=\nabla\cdot\left(\frac{\mathbf{e}_1}{h_2h_3}\right)$$ where $$h_i = h_i (u_1,u_2,u_3)$$ ? I don't understand this. It looks like we get something like $$\begin{pmatrix}f_1\dfrac{\partial}{\partial u_1}\\f_2\dfrac{\partial}{\partial u_2} \\ f_3\dfrac{\partial}{\partial u_3}\end{pmatrix}\cdot \begin{pmatrix}\dfrac{1}{h_2h_3}\\0\\0\end{pmatrix}=f_1\frac{\partial}{\partial u_1}\left(\frac{1}{h_2h_3}\right) = 0$$ where $$f_1 = f_1(u_1,u_2,u_3)$$ is some function not yet derived. Note that this problem is part of proving the form of the divergence operator as shown in "Proposition 7.1 ii)", so the divergence operator has not yet been confirmed, but I guess one can assume the form shown... or can we ?

BTW, please don't answer this question by deriving the answer for $$\nabla\cdot(A_1\mathbf{e}_1)$$ using tensors, unless it is shown why $$(A_1h_2h_3)\nabla\cdot (\nabla u_2\times \nabla u_3)=0$$

EDIT:===============================================================

I don't want to answer my own question, so I rather edit it...

You can use the identity $$\nabla\cdot(\mathbf{A}\times\mathbf{B})=\mathbf{B}\cdot(\nabla\times\mathbf{A})-\mathbf{A}\cdot(\nabla\times\mathbf{B})$$ to get $$\nabla\cdot(\nabla u_2\times\nabla u_3)=\nabla u_3\cdot(\nabla\times\nabla u_2)-\nabla u_2\cdot(\nabla \times \nabla u_2)$$ and then $$\nabla\times\nabla \psi=\mathbf{0}$$ That is, the curl of a gradient is always zero... So that explains the original question, but where am I wrong in the original derivation, leading to $$\nabla\cdot\left(\frac{\mathbf{e}_1}{h_2h_3}\right)$$