homework and exercises - Why are EM waves transverse?

I was reading Griffiths' Introduction to Electrodynamics, specifically the section on plane waves. I can see that if we want a transverse wave traveling in the $z$ direction that we are only going to want our waves to have $x$ and $y$ components, but the reasoning in Griffiths' left me confused.

We start with electric and magnetic field waves of the form

$E(z,t) = E_{0}e^{i(kz-\omega t)}$

$B(z,t) = B_{0}e^{i(kz-\omega t)}$

Since we are in free space, we have that $\nabla \cdot E = \nabla \cdot B = 0$.

Now comes the crucial step: Griffiths claims that these two facts immediately imply that

$(E_{0})_{z} = (B_{0})_{z} = 0$

I wasn't sure how this followed. I know that if I want my waves to be planar, that I need the x and y derivates of the fields to be 0, so that I have a constant magnitude over a front of constant phase, but I wasn't sure how to see that z derivative had to be zero as well. It seems that if you had an electric field plane wave whose real part was varying in space as a sine function, that if you were to look at its z derivative that you would get a cosine function.

Answer

Let's take a slightly more general case: Consider a wave with wave vector $\vec k=(k_x,k_y,k_z)$, with the electric field given by $$\vec E=\vec E_0\ e^{i(\vec k \cdot \vec r-\omega t)} $$ where $\vec r=(x,y,z)$. Now, we wan't to satisfy Maxwell's equations in the vacuum, including Gauss' law: $$\vec \nabla \cdot \vec E=0$$ The derivative is quite easily evaluated explicitly $$ \vec \nabla\cdot \vec E=\vec \nabla \cdot \bigl(\vec E_0\ e^{i(\vec k \cdot \vec r-\omega t)}\bigr)=i\vec k \cdot \vec E_0 e^{i(\vec k \cdot \vec r -\omega t)} $$

In order to satisfy Gauss' law, we must impose: $$\vec k \cdot \vec E_0=\ \text{?}$$

Physically, this means that the direction of propagation is always $\dots$ to the electric field. The exact same argument applies for the $\vec B$-field.

I leave it as an exercise to the reader to convince him(/her/it)self that the question as originally posed is equivalent, i.e. that we can assume without loss of generality that $\vec k = (0,0,k_z)$, resulting in the conclusion reached by Griffiths.