The Laplacian operator, $\nabla ^2$ can be defined to operate on both scalar fields and vector fields. If $f$ is a scalar field, then
$$\bbox[5px,border:2px solid #C0A000]{\nabla ^2 f(\vec r)\equiv \nabla \cdot \nabla f(\vec r)}$$
which in Cartesian coordinates can be written
$$\begin{align}
\nabla \cdot \nabla f(\vec r)&=\sum_{i=1}^{3}\sum_{j=1}^{3}\left(\hat x_i\frac{\partial}{\partial x_i}\right)\cdot \hat x_j\frac{\partial f(\vec r)}{\partial x_j}\\\\
&=\sum_{i=1}^{3}\frac{\partial^2f(\vec r)}{\partial x_i^2}
\end{align}$$
If $\vec f(\vec r)$ is a vector field, then the
$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 \vec f(\vec r)\equiv \nabla \left(\nabla \cdot \vec f(\vec r)\right)-\nabla \times \nabla \times \vec f(\vec r)}$$
which in Cartesian coordinates can be written
$$\begin{align}
\nabla (\nabla \cdot \vec f(\vec r)-\nabla \times \nabla \times \vec f(\vec r)
&=\sum_{i=1}^{3}\sum_{j=1}^{3}\hat x_i\frac{\partial}{\partial x_i}\frac{\partial f_j(\vec r)}{\partial x_j}-\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{k=1}^{3}\left(\hat x_i\times(\hat x_j\times \hat x_k)\frac{\partial^2 f_k(\vec r)}{\partial x_i \partial x_j}\right) \\\\
&=\sum_{i=1}^{3}\frac{\partial^2 \vec f(\vec r)}{\partial x_i^2}
\end{align}$$
We note that while in Cartesian coordinates, the forms for the Laplacian on a scalar and a vector are identical, this is not the case in other curvilinear coordinates.
You want to prove $\nabla\times X=0$ with $X:=(v\cdot\nabla)v+v\times(\nabla\times v)$, i.e.$$X_i=v_j\partial_jv_i+\underbrace{\epsilon_{ijk}\epsilon_{klm}}_{\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}}v_j\partial_lv_m=v_j\partial_jv_i+v_j\partial_iv_j-v_j\partial_jv_i=v_j\partial_iv_j=\partial_i(\tfrac12v_j^2).$$In other words, $X=\nabla(\tfrac12v^2)$, making the result trivial by @MarkViola's hint.
Best Answer
By definition,
$\nabla^2u = \nabla \cdot \nabla u \tag{1}$
for any sufficiently differentiable function $u$; I guess taking $u \in C^2(R^n, R)$ "suffices". Then
$\nabla^2(fg) = \nabla \cdot \nabla (fg), \tag{2}$
and by the product (Leibniz) rule for derivatives, which can easily seen to apply to gradients since in fact
$\dfrac{\partial (fg)}{\partial s} = \dfrac{\partial f}{\partial s}g + f \dfrac{\partial g}{\partial s}, \tag{3}$
which holds for any variable $s$ upon which $f$ and $g$ may depend, we have
$\nabla(fg) = f(\nabla g) + g(\nabla f). \tag{4}$
Next, we use the formula
$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X, \tag{5}$
which holds for a differentiable function $f$ and vector field $X$. It too is a variant of the Leibniz rule, and may easily be verified by working out the relevant expressions in terms of coordinates. See also this wikipedia entry. Using (4) and (5) together yields
$\nabla \cdot \nabla (fg) = \nabla \cdot (f \nabla g + g\nabla f) = \nabla f \cdot \nabla g + f\nabla^2g + \nabla g \cdot \nabla f + g \nabla^2f, \tag{6}$
and bringing it all together yields
$\nabla^2(fg) = f \nabla^2g + g\nabla^2f + 2 \nabla f \cdot \nabla g. \tag{7}$
Hope this helps. Cheers, and
Fiat Lux!!!