Prove that $\nabla \times \vec F =0 \implies \vec F = \nabla f$ How do I prove that if $\nabla \times \vec F = \vec 0 $ then $\vec F = \nabla f$ for some scalar field $f$?
My lecturer only proved the converse, which follows easily from the symmetry of mixed partial derivatives.
Preferably I would like a simple method of proof, not something powerful like Stokes theorem.
 A: Assume your vector field satisfies $\vec{\nabla}\times\vec{F} = \vec{0}$ on some open
star-shaped region with respect to the origin. i.e. if $x$ belongs to the region, so does the line segment joining $\vec{x}$ and the origin. Define 
$$f(\vec{x}) = \int_0^1 \vec{x} \cdot \vec{F}(t\vec{x}) dt$$
For any fixed unit vector $\hat{n}$, we have
$$\begin{align}
\hat{n}\cdot \vec{\nabla} f(\vec{x}) =
\left.\frac{d}{ds}f(\vec{x} + s\hat{n})\right|_{s=0}
&= \left.\frac{d}{ds}\left[\int_0^1 (\vec{x} + s\hat{n})\cdot \vec{F}(t(\vec{x} + s\hat{n})) dt\right] \right|_{s=0}\\
&= \int_0^1 \left( \hat{n}\cdot \vec{F}(t\vec{x}) + t \sum_{a,b=1}^3 x_a n_b \left.\frac{\partial F_a(\vec{z})}{\partial z_b} \right|_{\vec{z}=t\vec{x}}  \right) dt\\
\bbox[4pt,border:1px solid blue]{
\vec{\nabla}\times\vec{F} = \vec{0} \implies
\frac{\partial F_a}{\partial z_b}
= \frac{\partial F_b}{\partial z_a}}\!\!\color{blue}{\rightarrow}
&= \int_0^1 \left( \hat{n}\cdot \vec{F}(t\vec{x}) + t \sum_{a,b=1}^3 x_a n_b \left.\frac{\partial F_b(\vec{z})}{\partial z_a} \right|_{\vec{z}=t\vec{x}}  \right) dt\\
&= \int_0^1 \left( \hat{n}\cdot \vec{F}(t\vec{x}) + t \frac{d}{dt}\left[\hat{n}\cdot \vec{F}(t\vec{x})\right] \right) dt\\
&= \int_0^1 \frac{d}{dt}\left[ t \hat{n}\cdot \vec{F}(t\vec{x})\right] dt\\
&= \hat{n}\cdot \vec{F}(\vec{x})
\end{align}
$$
Since the direction of $\hat{n}$ is arbitrary, this leads to
$\vec{\nabla}f(\vec{x}) = \vec{F}(\vec{x})$ as desired.
A: Disclaimer: this answer is somewhat "sophisticated", and requires some knowledge of algebraic topology. 
Let $X$ be a simply connected $C^\infty$-manifold. Then we have for the de Rham cohomology
$H^1_{dR}(X)=H^1(X;\mathbf{R})=\mathrm{Hom}(H_1(X),\mathbf{R})=\mathrm{Hom}(\pi_1(X)^{\mathrm{ab}},\mathbf{R})=0$ (using de Rham's theorem, the universal coefficient theorem, and Hurewitz's theorem) since $\pi_1(X)=0$. Thus every closed $1$-form is exact.
Now if you take $X$ to be a simply connected open subset of $\mathbf{R}^3$, and if $(f,g,h)$ is your vector field on $X$, define the $1$-form $\omega=fdx+gdy+hdz\in\Omega^1(X)$. Saying that the curl of $(f,g,h)$ vanishes amounts to $\star d\omega=0$; but $d\omega=\star\star d\omega=0$, so that by the above result there is $\phi\in C^\infty(X)=\Omega^0(X)$ with $\omega=d\phi$. But this amounts to $\nabla\phi=(f,g,h)$.
