The gradient operator is defined as \[\nabla = \left\langle \frac{\partial }{\partial x},\frac{\partial }{\partial y}, \frac{\partial }{\partial z} \right\rangle\]
Assume the vector field \(\vec{F}\) is conservative. Therefore, \(\vec{F} = \nabla f\) for some function \(f\) in \(\mathbb{R}^3\). Take the curl of \(\vec{F}\):
\[\begin{aligned} \nabla \times \vec{F} &= \nabla \times \nabla f \\[1ex] &= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\\\end{vmatrix} \\[1ex] &= \left\langle \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial^2 f}{\partial z \partial y}, -\left(\frac{\partial^2 f}{\partial x \partial z} - \frac{\partial^2 f}{\partial z \partial x}\right), \frac{\partial^2 f}{\partial x \partial y} - \frac{\partial^2 f}{\partial y \partial x} \right\rangle \label{curlvector} \end{aligned}\]
Since \(f\) is the potential function of some conservative vector field,
\[\frac{\partial^2 f}{\partial y \partial z} = \frac{\partial^2 f}{\partial z \partial y},\quad \frac{\partial^2 f}{\partial x \partial z} = \frac{\partial^2 f}{\partial z \partial x},\quad \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}\]
As such, Equation [curlvector] reduces to
\[\nabla \times \vec{F} = \vec{0}\] Therefore, if \(\vec{F}\) is a conservative vector field, then the curl of \(\vec{F}\) is \(\vec{0}\).
Assume \(\nabla \times \vec{F} = \vec{0}\) for some vector-valued function \(\vec{F}(x,y,z) \in \mathbb{R}^3,\; x,y,z \in \mathbb{R}\)
Assume \(x \in \mathbb{R}\).
Note: a feature of \(\mathbb{R}\) is that \(\exists ! x \in \mathbb{R}(x = x)\)
Thusly, \(x=x\)
QED \(\blacksquare\)
This proof is left to the reader as an excercise in futility.