= Solution
In <Einstein summation convention>, the <curl> is
$$
(\nabla\times\mathbf V)_i=\epsilon_{ijk}\partial_jV_k.
$$
Apply another <curl> and use the <contraction of two Levi-Civita symbols>:
$$
\begin{aligned}(\nabla\times(\nabla\times\mathbf V))_i&=\epsilon_{ijk}\epsilon_{klm}\partial_j\partial_lV_m\\&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j\partial_lV_m\\&=\partial_i(\partial_jV_j)-\partial_j\partial_jV_i.\end{aligned}
$$
Smoothness allows the <partial derivatives> to commute. \b[This proves the <curl of the curl identity>:]
$$
\boxed{\nabla\times(\nabla\times\mathbf V)=\nabla(\nabla\cdot\mathbf V)-\nabla^2\mathbf V.}
$$
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