= Solution
The <four-gradient> rules from <1/a> give
$$
\partial'_\mu\partial'^\mu
=(\Lambda^{-1})^\nu{}_{\mu}\partial_\nu
\Lambda^\mu{}_{\rho}\partial^\rho
=\delta^\nu{}_{\rho}\partial_\nu\partial^\rho
=\partial_\nu\partial^\nu.
$$
The <Lorentz transformation> coefficients are constant, so the <partial derivatives> do not differentiate them. This proves the <Lorentz invariance> of the <d'Alembert operator>:
$$
\boxed{\Box=\partial_\mu\partial^\mu=\nabla^2-c^{-2}\partial_t^2.}
$$
For example, the two longitudinal terms transform as $-\gamma^2(\partial_0+\beta\partial_1)^2+\gamma^2(\partial_1+\beta\partial_0)^2=-\partial_0^2+\partial_1^2$, using $\gamma^2(1-\beta^2)=1$ and commuting <partial derivatives>. The transverse terms are unchanged. Applied to a <scalar field>, $\Box$ therefore gives a <Lorentz scalar>.
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