Assignment of Subscripts in Einstein Summation Notation I'm trying to understand the following conversion from vector form into Einstein summation notation, found on P2 of http://www.stanford.edu/~vkl/research/notes/index_not.pdf which states:

Show $\mathbf{v} \cdot \nabla\mathbf{v} = \nabla\left(\frac{|\mathbf{v}|^2}{2}\right)+(\color{brown}{\nabla \times \mathbf{v})} \times \mathbf{v}$
Proof: $v_a \partial_a v_b = \partial_b\left(\frac{v_av_a}{2} \right) + \epsilon_{bac}\color{brown}{(\epsilon_{adf}\partial_dv_f)}v_c \tag{*}$ (Rest of proof omitted here)

$\Large{\text{Question #1.}}$How did they get the LHS of (*)? I don't think my course covers the gradient of a vector, so I don't know how to convert it into Einstein notation. Or is it supposed to be $\nabla \cdot \mathbf{v}$?
$\Large{\text{Question #2.}} $ The solution seems to be working with the $b$th component of $ \mathbf{v} \cdot \nabla\mathbf{v}$, but it doesn't say so. Because $\mathbf{u} \times \mathbf{v} = \epsilon_{acb}u_av_j\color{red}{\mathbf{\hat{e_b}}}$,
is the solution missing  $\color{red}{\mathbf{\hat{e_b}}}$ on the RHS:  $$\partial_b\left(\frac{v_av_a}{2} \right) + \underbrace{\epsilon_{acb}}_{\Large{= \epsilon_{bac}}}\color{brown}{(\epsilon_{adf}\partial_dv_f)}v_c\color{red}{\mathbf{\hat{e_b}} }?$$

Here are two supplementary questions in response to tom's answer:
$\Large{\text{Question #1.1.}} $ How did you realise that $\mathbf{v} \cdot \nabla\mathbf{v}$ should've been written as $(\mathbf{v} \cdot \nabla)\mathbf{v}$?
$\Large{\text{Question #2.1.}}$ I don't understand your answer. $((\color{brown}{\nabla \times \mathbf{v})} \times \mathbf{v})$is a vector so why does the given solution not convert this to $\epsilon_{acb}\color{brown}{(\epsilon_{adf}\partial_dv_f)}v_c\color{red}{\mathbf{\hat{e_b}} }$?

Here are two supplementary questions in response to Muphrid's answer:
$\Large{\text{Question #1.2.}} $ What would be the "order of operations" which you mentioned in your answer?
$\Large{\text{Question #2.2.}}$ Sorry, I don't understand what you mean by "...it would otherwise appear as the same basis vector in each term." Could you please clarify?

Here are two supplementary questions in response to Muphrid's 2nd answer:
$\Large{\text{Question #2.3.}}$ Could you please explain how a free index ($\color{red}{b}$ here) means that we are looking at the component of this free index ($\color{red}{b}$th component here)?
 A: Q1:
Well $v\cdot \nabla v$ should be writen in form $(v \cdot \nabla) v$. 
Nabla is $$\nabla = (\partial_x,\partial_y,\partial_z).$$ So 
$$v\cdot \nabla = v_x\partial_x+v_y\partial_y+v_z\partial_z$$
Of course you can define gradient of vector valued function see Jacobi matrix
Q2:
Well no it is not. If by $\hat{e}_b$ you mean b-th basis vector.
$\partial_b\left(\frac{v_av_a}{2} \right)$ is scalar so 
$\epsilon_{bac}{(\epsilon_{adf}\partial_dv_f)}v_c $ has to be scalar too
Q1.1: Don't know what to say, but may I ask what is your definition of $\textbf{v} \cdot \nabla \textbf{v}$ ?
For me it is $\textbf{v} \cdot \nabla \textbf{v} =  (v_x\partial_x+v_y\partial_y+v_z\partial_z)\textbf{v}$ so there is not much to talk about.
Q2.1:
Yes you are right that $\mathbf{u} \times \mathbf{v} = \epsilon_{abc}u_av_b\mathbf{\hat{e}_c}$ the same is with $\mathbf{v} = v_a \mathbf{\hat{e}_a}$ and $\nabla = \mathbf{\hat{e}_a} \partial_a $.
So  $\nabla\left(\frac{|\mathbf{v}|^2}{2}\right) = \frac{1}{2}\mathbf{\hat{e}_a}\partial_a (v_bv_b)$ and $(\nabla \times \mathbf{v}) \times \mathbf{v} = \epsilon_{abc} (\epsilon_{bef} \partial_e v_f) v_c \mathbf{\hat{e}_c}$.
For more info on term $v\cdot \nabla v$ see Convective acceleration in Navier-Stokes equations
A: (1) It's $(v \cdot \nabla) v$, the directional derivative.  This is a common notation for it (dropping the brackets) because the gradient of a vector field doesn't fall under the scope of vector calculus--there is nothing else it could be.  You could also consider it an order of operations sort of thing, if you like.
(2) No, $\hat e_b$ is not missing, or rather, if it is missing, it is "missing" from every term.  The basis vector has been dropped from both sides for convenience.  You can verify that $b$ is the only free index on both sides, and it would otherwise appear as the same basis vector in each term.
