We wish to show that if $\nabla \times \mb{F} = 0$ in some open set $U$ and for some differentiable field, $\mb{F}$ in $U$, then there exists a potential, $\phi$, such that $\nabla \phi = 0$ **as long as** $U$ is simply connected. We create this explicitly. Let
\[
\phi(\mb{x}) = \int_{\mb{x}_0}^{\mb{x}} \mb{F} \cdot \de\mb{r'}
\]
where the integration begins from $\mb{x}_0$ and ends at $\mb{x}$ along a contour lying within $U$. By Stokes' Theorem, the contour is irrelevant since for any two different contours, $\int_{I_1} + \int_{-I_2} \mb{F} \, \de{\mb{r}'} = \iint_S \nabla \times \mb{F} \, \de{\mb{S}} = 0$. Notice that this statement is invalidated if $U$ is not simply connected. For instance, if $U$ is an annulus, then we must add both inner and outer contours.

We must now only demonstrate $\nabla \phi = \mb{F}$, but it is strange to speak about $\nabla \phi$ given the definition of $\phi$ as an integral. Let $\mb{v}$ be an arbitrary unit vector. If we can show that \[ \nabla \phi \cdot \mb{v} = \mb{F} \cdot \mb{v}, \]

then we are done. However, the left hand side is simply the directional derivative (rate of change) of $\phi$ at $\mb{x}$ moving in the direction $\mb{v}$. By definition, this is \[ \dd{\phi}{s} = \lim_{s \to 0} \frac{\phi(\mb{x} + s\mb{v}) - \phi(\mb{x})}{s} = \lim_{s \to 0} \frac{1}{s} \int_{\mb{x}}^{\mb{x} + s\mb{v}} \mb{F} \cdot \de{\mb{r}'} \]

where the path of integration is the straight line $\mb{r}' = \mb{x} + s'\mb{u}$, where s' is between 0 and s. Thus we have \[ \dd{\phi}{s} = \lim_{s \to 0} \frac{1}{s} \int_{0}^{s} \mb{F} \cdot \mb{v} \, \de{s'} = \lim_{s \to 0} \frac{(\mb{F}(\mb{x}) \cdot \mb{v}) s}{s} = \mb{F}(\mb{x}) \cdot \mb{v}. \]

where the second equality comes from a Taylor expansion about $s = 0$. Thus $\nabla \phi \cdot \mb{v} = \mb{F} \cdot \mb{v}$ for arbitrary unit vector (and hence arbitrary vector). QED.