Yes, $\star(\text{d}\omega)$ is one definition to $\text{rot} \ v$. I need to prove that if $\text{rot} \ v=0$ then $v$ is locally exact (derives locally from a potential, more context can be provided here). And yes, I've checked others answers, like this one (and I know how to solve this problem). But this came to my mind, because in this case, as a consequence, $\text{d}\omega = 0$. Is this true in general?
The definition I have for the Hodge Star Operation on a $k-$form of $\Bbb{R}^{n}$ is that \begin{align*} \star(d_{x_{i_{1}}} \wedge \dots \wedge d_{x_{i_{k}}}) = (-1)^{\text{sgn}\ \sigma} (d_{x_{j_{1}}} \wedge \dots \wedge d_{x_{j_{n-k}}}),\ \text{where} \ (i_{1}\dots i_{k}j_{i}\dots j_{n-k}) \in S_{n}. \end{align*} If the latter is $0$, then $d_{x_{j_{a}}}=d_{x_{j_{b}}}$ for some $a,b$. But all of $d_{x}$'s form a base for the dual of $k-$form of $\Bbb{R}^{n}$, so this can't happen. This is enough to conclude that? Thanks in advance!