The key fact is that gradient fields are conservative, and conservative fields have zero circulation, which is exactly the curl being zero.
Start from $\vec{A} = \vec{\nabla}\phi$. The component of curl about any axis measures how much the field circulates. For a gradient field, moving around any closed loop returns the scalar $\phi$ to its starting value, so the net line integral $\oint \vec{A}\cdot d\vec{l} = 0$ for every loop.
By Stokes' theorem, $\oint \vec{A}\cdot d\vec{l} = \int (\vec{\nabla}\times\vec{A})\cdot d\vec{S}$. Since the loop integral vanishes for all loops, the integrand must vanish everywhere, so $\vec{\nabla}\times\vec{A} = 0$.
Checking the other choices: $\vec{\nabla}\cdot\vec{A}=0$ is the condition for a field with no sources (solenoidal), not for a gradient. The remaining options are not standard conditions for a gradient field. Hence the correct condition is that the curl vanishes.
\[\boxed{\vec{\nabla} \times \vec{A} = 0}\]