Why is $(\mathbf{v} \cdot \nabla)\mathbf{v} = (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)$? The Convective Derivative or Material Derivative is usually written as $\frac{D}{Dt}=\frac{\partial}{\partial t} + \mathbf{v} \cdot \nabla$.  According to MathWorld, this equation, multiplied with ${\bf{v}}$ equals:
$$
    \frac{D \mathbf{v}}{Dt} = \frac{\partial \mathbf{v}}{\partial t} + (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)
$$
Clearly, it must hold that;
$$
    (\mathbf{v} \cdot \nabla)\mathbf{v} = (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)
$$
However, I do not spot why this is true. What is the (trivial) identity that I am missing?

We can write expand the components using Levi-Civita symbol (using the convention of summing repeated indices):
$$
\left[(\nabla \times \mathbf{v})\right]_i = \epsilon_{ijk} \left(\partial_j v_k\right).
$$
$$
\left[(\nabla \times \mathbf{v}) \times \mathbf{v}\right]_{l} = 
\epsilon_{lim}  \left[\nabla \times \mathbf{v} \right]_{i} v_m=
\epsilon_{lim} \epsilon_{ijk} v_m \left(\partial_j v_k\right) = -\epsilon_{ilm} \epsilon_{ijk} v_m \left(\partial_j v_k\right)
$$
We also have this identity between Levi-Civita and Kronecker delta:
$$
\epsilon_{ilm} \epsilon_{ijk} = \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj}.
$$
By plugging this identity in the second equation:
$$
\begin{align}
\left[(\nabla \times \mathbf{v}) \times \mathbf{v}\right]_{l} &=
- \delta_{lj} \delta_{mk}  v_m \left(\partial_j v_k\right) + \delta_{lk} \delta_{mj} v_m \left(\partial_j v_k\right)\\ 
&= - v_k \partial_l v_k + v_j \partial_j v_l \\
&= - \frac{1}{2} \partial_l (v_k v_k) + (v_j \partial_j) v_l
\\
&= - \left[\nabla (\frac{1}{2} \mathbf{v}^2)\right]_l + 
\left[(\mathbf{v} \cdot \nabla)\mathbf{v}\right]_l
\end{align}
$$
If you are not familiar with Levi-Civita symbol, expand $(\nabla \times \mathbf{v}) \times \mathbf{v}$ for its $x$ component. It will be similar for other components.
