Obviously, if you have spatially constant vector fields, their spatial rates of change will be 0 in any direction. Application of the divergence and curl produce 0 or the zero vector, respectively. Sort of like how a constant function $f(x)=c$ in single variable calculus has a derivative of 0 everywhere.
Added: Using x,y,z to represent the standard Cartesian basis vectors instead of $\hat{i},\hat{j},\hat{k}$ is a sure recipe for trouble. How do you represent the scalar field $f(x,y,z)=3.5x + 6y + 4z$?
Addendum 2: Note that for coordinate systems in which the basis vectors are a function of space (e.g. cylindrical)
$$
\hat{e}_r = \cos(\theta) \hat{e}_1 + \sin(\theta) \hat{e}_2 \\
\hat{e}_\theta = -sin(\theta) \hat{e}_1 + \cos(\theta) \hat{e}_2 \\
\hat{e}_z = \hat{e}_3
$$
If you have a vector field with (cylindrical) components $5,6,3$ everywhere, it is not really a constant vector field at all because the basis vectors change from point to point
(Notice the contrast with rectangular cartesian, where a vector field with components $5,6, 3$ everywhere is a constant vector field.)
In fact, if you transform the components of this vector back to cartesian, the vector $5,6,3$ has cartesian components which are functions of space. E.g. its x component is $5 \cos(\theta) \hat{e}_1 - 6 \sin(\theta) \hat{e}_2 + \hat{e}_3$, where $\theta=\tan^{-1}(y/x)$
Best Answer
Remember that in the analogous case $\nabla \times \nabla f = 0$, some intuition for the result can be attained by integration: by Green's theorem this is equivalent to $\int \nabla f \cdot ds = 0$ around every closed loop, which is true because $\int_{\gamma} \nabla f \cdot ds = f(\gamma(1)) - f(\gamma(0)).$ Thus our intuition is that curl measures circulation, and $\nabla f$ cannot circulate because this would introduce a discontinuity in $f$ around a loop.
Let's try the same thing: by the divergence theorem, it suffices to show that $\int_\Sigma (\nabla \times V) \cdot \hat n\ dA = 0$ for every closed surface $\Sigma$. By Stokes' theorem we know $$\int_\Sigma (\nabla \times V) \cdot \hat n\ dA = \int_{\partial \Sigma}V\cdot ds,$$which vanishes because $\Sigma$ is closed (i.e. $\partial \Sigma = \emptyset$).
In more intuitive terms, the divergence measures flux through a small cube; but the flux of a curl through a closed surface must be zero because there is no boundary curve for the circulation to accumulate upon.
As Nameless alluded to in his comment, you can get a more unified understanding of what's going on here by studying differential forms. All these geometric differential operators $\nabla, \nabla \times, \nabla \cdot$ are exterior derivatives $d_0,d_1,d_2$, and the identity $d_{k+1} \circ d_k = 0$ can be seen either by expanding out the partial derivative expression and noting that everything cancels (what I assume you've done in your proof), but also by applying the general Stokes theorem twice and noting that the boundary of a boundary is always empty.