Does taking $\nabla\times$ infinity times from an arbitrary vector exists? Is it possible to get the value of:
\begin{equation}
\underbrace{\left[\nabla\times\left[\nabla\times\left[\ldots\nabla\times\right.\right.\right.}_{\infty\text{-times taking curl operator}}\mathbf{V}\left.\left.\left.\right]\right]\ldots\right] = ?
\end{equation}
For any possible values of vector $\mathbf{V}$.

The identity
$$
\nabla \times (\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla^2 \mathbf{A}
$$
is standard (where $\nabla^2$ denotes the component-wise Laplacian). Applying it repeatedly to $\nabla \times \mathbf{A}$ and using the fact that $\nabla \times (\nabla f)$ vanishes, we can see inductively that
$$
(\nabla \times)^{2n+1}\mathbf{A}=(-1)^n\nabla \times (\nabla^2)^n \mathbf{A} \, .
$$
So, when there's any hope of your thing converging, it'll be because the iterations of $-\nabla^2$ converge component-wise. Roughly speaking, this will happen when the Fourier transforms of your components are supported within some appropriate sphere, but getting the details right could be tricky (especially in cases where it includes some piece of the sphere's boundary).
A little more precisely, the equivalent of $-\nabla^2$ in the frequency domain is multiplication by $4\pi^2|\xi|^2$, so if $\operatorname{supp} \hat f \subset \{|\xi|<1/(2\pi)\}$ everything will converge to $0$. If $\operatorname{supp} \hat f \subset \{|\xi|\leq 1/(2\pi)\}$ everything should still converge; if $\hat f$ is a distribution with positive mass on the boundary of the sphere it should converge to that, but the analysis is icky enough that I don't want to say anything too definite...
You then have to worry about the even iterates, which you can get by replacing $\mathbf{A}$ with $\nabla \times\mathbf{A}$ everywhere in the above identity; there's no real reason why the even and odd iterates ought to have much to do with each other. On the other hand, in the nice case everything will converge to $0$ anyway, so this won't really matter.
