Divergence of a Curl Proof Not coming out to 0 In all physics courses we are taught that the divergence of a curl is always zero: $\nabla \cdot(\nabla \times\vec{V}) = 0$
So to prove this to myself I simply solve it to get $0$, but I am not coming up to zero. Can someone please point out where my mistake is?
\begin{align*}
\nabla \times \vec{V}&= \begin{vmatrix}\hat{x} &\hat{y} &\hat{z} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 
V_x & V_y & V_z \end{vmatrix}\\
&=\left( \frac{\partial}{\partial y}V_z -  \frac{\partial}{\partial z}V_y\right)\hat{x} + \left( \frac{\partial}{\partial x}V_z -  \frac{\partial}{\partial z}V_x\right)\hat{y} + \left( \frac{\partial}{\partial x}V_y -  \frac{\partial}{\partial y}V_x\right)\hat{z}\\
&=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}V_z -  \frac{\partial}{\partial z}V_y\right)\hat{x} + \frac{\partial}{\partial y}\left( \frac{\partial}{\partial x}V_z -  \frac{\partial}{\partial z}V_x\right)\hat{y} + \frac{\partial}{\partial z}\left( \frac{\partial}{\partial x}V_y -  \frac{\partial}{\partial y}V_x\right)\hat{z}\\
&=
\frac{\partial^2 V_z}{\partial x \partial y} - 
\frac{\partial^2 V_y}{\partial x \partial z} + 
\frac{\partial^2 V_z}{\partial y \partial x} - 
\frac{\partial^2 V_x}{\partial y \partial z} + 
\frac{\partial^2 V_y}{\partial x \partial z} - 
\frac{\partial^2 V_x}{\partial z \partial y}
\end{align*}
Only one term cancels out the $\frac{\partial^2 V_y}{\partial x \partial z}$?
Leaving me with:
$$
\nabla\cdot(\nabla \times\vec{V}) =
2\left (\frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_x}{\partial z \partial y}\right ).
$$
I know I have screwed up somewhere, but I have been staring at my work for a while now and I can't seem to figure out where I went wrong. All terms should cancel out...
Thank you.
 A: You have a sign error in your middle term in $\nabla \times \vec{V}$ (the coefficient of $\hat{y}$).  If you're calculating the determinant by expansion by minors, remember that the signs of the terms need to alternate;  see eq. (2) in the above link.
A: The correct form for curl is
$$
\nabla \times \vec{V}=\left( \frac{\partial}{\partial y}V_z -  \frac{\partial}{\partial z}V_y\right)\hat{x} + \underbrace{\left(\frac{\partial}{\partial z}V_x - \frac{\partial}{\partial x}V_z\right)}_{\text{note the switch}}\hat{y} + \left( \frac{\partial}{\partial x}V_y -  \frac{\partial}{\partial y}V_x\right)\hat{z}
$$
Another version of curl (which is convenient for this proof) is
$$
\nabla \times \vec{V} = \hat x \times \frac{\partial V}{\partial x} + \hat y \times \frac{\partial V}{\partial y} + 
\hat z \times \frac{\partial V}{\partial z}
$$
Similarly, divergence is given as
$$
\nabla \cdot \vec{V} = \hat x \cdot \frac{\partial V}{\partial x} + \hat y \cdot \frac{\partial V}{\partial y} + 
\hat z \cdot \frac{\partial V}{\partial z}
$$
