Proof of the vector calculus identity $\nabla\cdot(u\times v)=v\cdot(\nabla\times u)-u\cdot(\nabla\times v)$ $$\nabla\cdot(u\times v)=v\cdot(\nabla\times u)-u\cdot(\nabla\times v)$$
How can we go about proving the above vector calculus identity component wise?
 A: You can do it the hard way (sit down and plug in two general vector fields, then slog through the calculations), or you can do it the smart way, namely by concentrating on two special cases: (1) $u=(f,0,0)$ and $v=(g,0,0)$ and (2) $u=(f,0,0)$ and $v=(0,g,0)$. That should be fairly easy.
Now notice that if you permute components cyclically, then the components of cross products and curls get permuted the same way, so from the two cases above you have now proved the formula for all vector fields $u$ and $v$ having only one nonzero component each.
Finally, note that any vector field is a sum of three vector fields of the type mentioned above, and since the desired formula is additive in $u$ and $v$ separately, you're done.
A: Alternatively, you could use suffix notation:
$$\begin{align*} \boldsymbol{\nabla}\cdot(\mathbf{u}\times\mathbf{v})
&=\partial_i\epsilon_{ijk}u_jv_k \quad\quad\text{definition of Levi-Civita symbol} \\
&=\epsilon_{ijk}\partial_iu_jv_k \quad\quad\text{Levi-Civita symbol is constant} \\
&=\epsilon_{ijk}(v_k\partial_iu_j+u_j\partial_iv_k) \quad\quad\text{derivative of a product} \\
&=\epsilon_{kij}v_k\partial_iu_j-\epsilon_{jik}u_j\partial_iv_k \quad\quad\text{distributivity and property of Levi-Civita symbol} \\
&=v_k\epsilon_{kij}\partial_iu_j-u_j\epsilon_{jik}\partial_iv_k \quad\quad\text{commutativity} \\
&=\mathbf{v}\cdot(\boldsymbol{\nabla}\times\mathbf{u})-
\mathbf{u}\cdot(\boldsymbol{\nabla}\times\mathbf{v}) \quad\quad\text{definition of Levi-Civita symbol}\end{align*}$$
