= Solution
Use <suffix notation>, summing repeated indices, and write $\partial_i=\partial/\partial x_i$. For a twice continuously differentiable <vector field>, mixed derivatives commute. Thus the <divergence> of its <curl> satisfies
$$
\partial_i(\epsilon_{ijk}\partial_jv_k)
=\epsilon_{ijk}\partial_i\partial_jv_k=0,
$$
because the derivatives are symmetric in $i,j$, while the <Levi-Civita symbol> is antisymmetric in those indices. Hence
$$
\boxed{\nabla\cdot(\nabla\times\mathbf v)=0}.
$$
For the second identity, use the <contraction of two Levi-Civita symbols>:
$$
\begin{aligned}
[\mathbf v\times(\nabla\times\mathbf v)]_i
&=\epsilon_{ijk}v_j\epsilon_{klm}\partial_lv_m\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})v_j\partial_lv_m\\
&=v_j\partial_iv_j-v_j\partial_jv_i\\
&=\partial_i\left(\tfrac12v_jv_j\right)-[(\mathbf v\cdot\nabla)\mathbf v]_i.
\end{aligned}
$$
Rearranging each component proves the <vorticity cross-product identity>
$$
\boxed{(\mathbf v\cdot\nabla)\mathbf v=
\nabla\left(\tfrac12|\mathbf v|^2\right)-\mathbf v\times(\nabla\times\mathbf v)}.
$$
The second identity needs first derivatives; the first additionally uses equality of the mixed second derivatives.
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