= Solution
For proper rotations the <Levi-Civita symbol> is invariant, so contracting the two rank-one tensors $\partial_j$ and $v_k$ gives $c_i=\epsilon_{ijk}\partial_jv_k$ with transformation $c_i'=R_{i\ell}c_\ell$. Thus the <curl> is a vector under proper rotations. Under an orientation-reversing orthogonal change it acquires an additional determinant sign: it is an <axial vector> when $v$ is a polar vector.
Commuting smooth partial derivatives and using epsilon contraction gives the <curl of the curl identity>:
$$
[\nabla\times(\nabla\times v)]_i=\epsilon_{ijk}\epsilon_{klm}\partial_j\partial_lv_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j\partial_lv_m=\partial_i(\partial_jv_j)-\partial_j\partial_jv_i.
$$
Now take the curl of <Faraday's law> and use the vacuum <Ampère-Maxwell equation>: $\nabla\times\nabla\times E=-\partial_t(\nabla\times B)=-c^{-2}E_{tt}$. Since $\nabla\cdot E=0$, the identity gives $\Delta E-c^{-2}E_{tt}=0$. Taking the curl of the <Ampère-Maxwell equation> gives the same result for $B$, using $\nabla\cdot B=0$. Hence both satisfy the <electromagnetic wave equation>
$$
\boxed{-\Delta E+c^{-2}E_{tt}=0,\qquad-\Delta B+c^{-2}B_{tt}=0.}
$$
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