Irrotational implies conservative without using the path-connection

Question

Let's consider a simply-connected domain $V$ in $\mathbb{R}^{3}$ and a smooth vector field $\mathbf{F}$ in $V$ (please don't answer considering other scenarios). Under this assumption, let's consider the following three statements

1. "$\mathbf{F}$ is conservative in $V$" (there exists a smooth scalar field $f$ in $V$ such that $\mathbf{F}=\nabla f$ everywhere)
2. "$\mathbf{F}$ is path-independent" (any line integral of $\mathbf{F}$ in $V$ depends on the extremes only)
3. "$\mathbf{F}$ is irrotational in $V$" ($\nabla\times\mathbf{F}=0$ in $V$)

It's known that:

• 1. is equivalent to 2.
• 2. is equivalent to 3.
• 1. (trivially) implies 3.

Usually, the statement 3. implies 1. is proved "climbing backwards": 3. implies 2. and 2. implies 1.

I was wondering if there exists any proof of 3. implies 1. without using the path independence. In a complete similar fashion I was wondering the same for divergence free vector fields, if they can be expressed in terms of the curl of another vector field.