Starting from Maxwell-equations in vacuum :
$$ \nabla \cdot \vec{E} = 0 $$ $$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} $$ $$ \nabla \cdot \vec{B} = 0 $$ $$ \nabla \times \vec{B} = \frac{1}{c^2}\frac{\partial \vec{E}}{\partial t} $$
We can show the existence of electromagnetic waves (using the identity $\nabla \times \nabla \times \vec{F} = \nabla(\nabla \cdot F) - \nabla^2 \vec{F})$ :
$$ \frac{\partial^2 \vec{E}}{\partial t^2} = c^2 \nabla^2 \vec{E} $$
$$ \frac{\partial^2 \vec{B}}{\partial t^2} = c^2 \nabla^2 \vec{B} $$
The solutions to these equations are the following for plane waves (using $\mathbb{C}$ notation) :
$$ \vec{E}(\vec{r}, t) = \vec{E_0}e^{i(\vec{k} \cdot \vec{r} - wt)} $$
$$ \vec{B}(\vec{r}, t) = \vec{B_0}e^{i(\vec{k} \cdot \vec{r} - wt)} $$
We can show (using divergence of the electric and magnetic field in vacuum) that these waves form an orthonormal basis $(\vec{E}, \vec{B}, \vec{k})$
However, I'm looking for a proof that :
$$ ||\vec{E}|| = c||\vec{B}|| $$
I've looked everywhere, in Griffith electrodynamics, in my books (Berkeley vol. II and III) but I've found nothing.