If $\vec F$ be defined in a simply connected region, then $\vec \nabla \times \vec F=0 $ $\implies$ $\exists$ a scalar potential $\phi$ such that $\nabla \phi=\vec F$. Prove it by using Helmholtz decomposition.
It is solved here in a different way Prove that $\nabla \times \vec F =0 \implies \vec F = \nabla f$
Using Helmholtz decomposition, we have:
$$F = \nabla \psi+ \nabla \times \vec G.$$
Then $$curl\, \vec F =curl(grad\,\psi)+\nabla\times(\nabla\times \vec G)$$ i.e, $$\vec 0=\vec 0+ \nabla\times(\nabla\times \vec G)\implies \nabla\times(\nabla\times \vec G)=\vec 0.$$
Now using, $\nabla\times(\nabla\times G) = \nabla (\nabla\cdot G) - \nabla^2 G$ or else, how to claim that $$F = \nabla \phi?$$
Please explain in details. Thank you.