Divergence of curl
Let \( \vec{A} \) be any vector.
Lets calculate curl of it:
\[ \nabla \times \vec{A} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
\frac{\partial}{ \partial x} & \frac{\partial}{ \partial y} & \frac{\partial}{ \partial z} \\
A_x & A_y & A_z
\end{vmatrix} =
\vec{i} ( \frac{ \partial A_z}{\partial y} - \frac{ \partial A_y}{\partial z} )+
\vec{j} ( \frac{ \partial A_x}{\partial z} - \frac{ \partial A_z}{\partial x} )+
\vec{k} ( \frac{ \partial A_y}{\partial x} - \frac{ \partial A_x}{\partial y} )
\]
It is a vector. Lets calculate divergence of it:
\[
\begin{multline}
\nabla \cdot ( \nabla \times \vec{A}) = \\
\frac{\partial}{\partial x}(\frac{ \partial A_z}{\partial y} - \frac{ \partial A_y}{\partial z}) +
\frac{\partial}{\partial y}(\frac{ \partial A_x}{\partial z} - \frac{ \partial A_z}{\partial x} ) +
\frac{\partial}{\partial z}( \frac{ \partial A_y}{\partial x} - \frac{ \partial A_x}{\partial y} )
\end{multline} = \\
\frac{ \partial^2 A_z}{ \partial x \partial y} - \frac{ \partial^2 A_y}{ \partial x \partial z} +
\frac{ \partial^2 A_x}{ \partial y \partial z} - \frac{ \partial^2 A_z}{ \partial y \partial x} +
\frac{ \partial^2 A_y}{ \partial z \partial x} - \frac{ \partial^2 A_x}{ \partial z \partial y} \tag{1}
\]
It is known that
\[ \frac{ \partial^2 f}{ \partial x \partial y} = \frac{ \partial^2 f}{ \partial y \partial x} \]
and (1) estimates to 0 for any \( \vec{A} \).