$\
ewcommand{\uvect}{\hat{\bf e}}$
Frist note that
$$ \
abla\cdot (\
abla\times{\bf E}) = 0 \tag{1} $$
You can show this is true by expanding each term (abusing a bit Einstein's notation):
$$ (\uvect_i\partial_i)\cdot(\uvect_k \epsilon_{klm}\partial_l E_m) = \delta_{ik}\epsilon_{klm}\partial_i\partial_l E_m = \epsilon_{klm}\partial_i\partial_l E_m = 0 $$
where I used the fact that $\epsilon_{klm} = -\epsilon_{lkm}$ whereas $\partial_k\partial_l E_m = \partial_l\partial_k E_m$. With this in mind note that
\begin{eqnarray} \
abla\times{\bf E} &=& \frac{\partial{\bf B}}{\partial t} \\\ \
abla\cdot(\
abla\times{\bf E}) &=& \
abla\cdot\left(\frac{\partial{\bf B}}{\partial t}\right) \\\ 0 &=& \frac{\partial}{\partial t}(\
abla \cdot {\bf B}) \tag{2} \end{eqnarray}