Vector Calculus - Curl of Vector I'm asked to prove the following identity, using index notation:
$(\nabla\times A)\times A=A \cdot\nabla A - \nabla(A \cdot A)$ 
However, when I work it out, I find that the actual solution should be: 
$(\nabla\times A)\times A=A \cdot\nabla A - \frac{1}{2}\nabla(A \cdot A)$ 
Am I missing something, or is the book wrong?

Expanding everything out componentwise, I get
$(\nabla\times A)\times A$
$$
\begin{align}
&{\lower{1pt}{\Large(}}(\partial_1,\partial_2,\partial_3)\times(A_1,A_2,A_3){\lower{1pt}{\Large)}}\times(A_1,A_2,A_3)\\
&=(\partial_2A_3-\partial_3A_2,\partial_3A_1-\partial_1A_3,\partial_1A_2-\partial_2A_1)\times(A_1,A_2,A_3)\\
&=(A_3\partial_3A_1+A_2\partial_2A_1-A_3\partial_1A_3-A_2\partial_1A_2,\\
&\hphantom{=(}A_1\partial_1A_2+A_3\partial_3A_2-A_1\partial_2A_1-A_3\partial_2A_3,\\
&\hphantom{=(}A_2\partial_2A_3+A_1\partial_1A_3-A_2\partial_3A_2-A_1\partial_3A_1)
\end{align}
$$
$(A\cdot\nabla)A$
$$
\begin{align}
&(A_1\partial_1+A_2\partial_2+A_3\partial_3)(A_1,A_2,A_3)\\
&=(A_1\partial_1A_1+A_2\partial_2A_1+A_3\partial_3A_1,\\
&\hphantom{=(}A_1\partial_1A_2+A_2\partial_2A_2+A_3\partial_3A_2,\\
&\hphantom{=(}A_1\partial_1A_3+A_2\partial_2A_3+A_3\partial_3A_3)
\end{align}
$$
$\frac12\nabla(A\cdot A)$
$$
\begin{align}
&\tfrac12(\partial_1,\partial_2,\partial_3)(A_1A_1+A_2A_2+A_3A_3)\\
&=(A_1\partial_1 A_1+A_2\partial_1 A_2+A_3\partial_1 A_3,\\
&\hphantom{=(}A_1\partial_2 A_1+A_2\partial_2 A_2+A_3\partial_2 A_3,\\
&\hphantom{=(}A_1\partial_3 A_1+A_2\partial_3 A_2+A_3\partial_3 A_3)
\end{align}
$$
And from these, I get
$$
(\nabla\times A)\times A=(A\cdot\nabla)A-\tfrac12\nabla(A\cdot A)
$$
