I want to show that: $$(n \cdot \nabla n)^2=(n \times \nabla \times n)^2$$ $n$ is a unit vector and $-n$ is essentially $n$ for this application. Using the vector triple product I can get the inner argument of the RHS to become:
$(n \times \nabla \times n)^2= [\nabla(n \cdot n)-n (\nabla \cdot n)]^2 =[-n (\nabla \cdot n)]^2=[n (\nabla \cdot n)]^2$
Using the definition of the dot product I get:
LHS = $[|n||\nabla n|cos\theta]^2=|\nabla n|^2cos^2\theta$ $$$$ RHS = $[n|\nabla||n|cos\theta]^2=n^2|\nabla|^2cos^2\theta$
So (if I'm on the right track of course) is $n^2|\nabla |^2=|\nabla n|^2$ necessarily?