Trying to prove Kelvin's Circulation Theorem, but struggling to see why the following equality holds:
$$\textbf{u} \cdot (d\textbf{l}\cdot \nabla)\textbf{u} = d\textbf{l} \cdot \nabla \left(\frac{1}{2} u^2 \right)$$
I don't really have much idea where to start. I can take that gradient on the RHS, but that is of a scalar function and I'm not sure how I'll be able to get the vector $\textbf{u}$ out in the way that looks anything like how it should on the LHS...
Any hints would be appreciated (don't mind a solution either)
Edit: added picture of $d\mathbf{l}$ representation from notes. 
\begin{align} d\textbf{l} \cdot \nabla \left(\frac{1}{2} u^2 \right) &\;=\; \sum_{\mu\, \nu} dl_{\mu} \, \frac{\partial}{\partial x_{\mu}} \, \frac{1}{2} u_{\nu}u_{\nu}\\ &\;=\; \sum_{\mu\, \nu} dl_{\mu} \, u_{\nu}\, \frac{\partial}{\partial x_{\mu}} u_{\nu}\\ &\;=\; \sum_{\mu\, \nu} u_{\nu} \, dl_{\mu} \, \frac{\partial}{\partial x_{\mu}} u_{\nu}\\ &\;=\; \mathbf{u} \cdot \bigl[ \left(d\mathbf{l}\cdot \mathbf{\nabla}\right) \mathbf{u} \bigr] \end{align}