In the equation you have, in the $(\nabla \times \ \textbf{v}) \times \textbf{v}$ term, there are parantheses around the first two terms, so it's not necessarily zero since first you do $\nabla \times \ \textbf{v}$, which is perpendicular to $\textbf{v}$, and then you cross it with $\textbf{v}$.
Starting with the vector triple product seems like a good approach. You can always write it out in tensor notation with the Levi-Civita symbol to get the $n^{th}$ component:
$$\left((\partial_i v_j) \epsilon_{ijk}\right) v_m \epsilon_{kmn},$$
where $\left((\partial_i v_j) \epsilon_{ijk}\right)$ is the $k^{th}$ component of $(\nabla \times \ \textbf{v})$.
Since $\epsilon_{kmn} = \epsilon_{mnk}$ and $\epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$, the $n^{th}$ component is equal to
$$ \begin{eqnarray}(\partial_i v_j) v_m \epsilon_{ijk}\epsilon_{mnk}\\
&=&(\partial_i v_j) v_m (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})\\
&=&(\partial_i v_j) v_m \delta_{im}\delta_{jn}-(\partial_i v_j) v_m\delta_{in}\delta_{jm}\\
&=& v_m \partial_m v_n -\partial_n v_m v_m.
\end{eqnarray}$$
Now, you can use what @PAM1499 noted (the chain rule) to write $\partial_n v_m v_m = \frac{1}{2} \partial_n (v_m v_m)$.
Putting everything back into vector form, you get:
$$(\nabla \times \ \textbf{v}) \times \textbf{v} = (\textbf{v} \cdot \nabla ) \textbf{v}- \frac{1}{2} \nabla(\textbf{v} \cdot \textbf{v}).$$
and all you need to do is rearrange the terms to get the identity.
