Proving zero identities in vector calculus with simple arguments involving index counting or symmetry? Consider the following table describing four second derivative operators.
             L   R
curl grad    2   1   vanishes identically
curl curl    3   1
div curl     3   0   vanishes identically
div grad     2   0

The idea of the $L$ and $R$ numbers is that they're counts of the number of indices you would expect to see on the left and right sides of an expression involving this operator, if it was expressed in index notation, with the following rules:


*

*Because indices can be contracted, all numbers are to be interpreted modulo 2, but in the above table I've just counted the raw number of indices without contraction.

*We ignore the fact that a curl requires a Levi-Civita symbol with three additional indices.
For example, if we take the equation $\operatorname{div}\operatorname{grad}f = g$ and translate it into index notation, we have
$$\nabla_i \nabla^i f = g.$$
There are two indices on the left, so $L=2$, and no indices on the right, $R=0$.
So in these four examples, I make the following observation: the derivative operator vanishes identically iff $L\ne R\ (\text{mod}\ 2)$. This rule would seem to make a lot of sense, and would probably convince me as a proof of the two zero identities, if the curl really was a tensor. But in reality the curl is not a tensor with one index, it's a tensor contracted with a Levi-Civita symbol, and because the Levi-Civita symbol has an odd number of indices in three dimensions, this messes up the parity of $L$.
I have two questions:
(1) Can this argument be patched up somehow?
(2) Can this type of argument be converted into a more elementary argument about symmetry, which could be presented in a way that was understandable to sophomore-level students who didn't know about tensors?
 A: This is not really a complete answer to my own question, only to part 2 for one particular zero identity, but it's too long for a comment, so I thought I'd post it here and see if it attracts any comments. I'm interested in whether it seems correct, and whether it can be simplified, clarified, or generalized to apply to more than one zero identity.
As a possible step toward the desired generalization, it's interesting to note that $\operatorname{curl}\operatorname{grad}$ and $\operatorname{div}\operatorname{curl}$ are identical when written in index notation. They differ only in that in $\operatorname{div}\operatorname{curl}$ we apply it to a field by contracting one index.
Argument that the divergence of a curl is zero
The kind of field that looks like it would have a nonvanishing value of this operation is something like $\textbf{F}=zx\hat{\textbf{y}}$, the idea being that we take a field like $x\hat{\textbf{y}}$ that has a curl in the $z$ direction, and then we give it some $z$ dependence so that there could be a divergence. In fact, any field that is differentiable near the origin can have its components approximated in that neighborhood by a function of this general form (a second-order mixed polynomial), so if we can prove that $\operatorname{div}(\operatorname{curl}\textbf{F})=0$ for this particular $\textbf{F}$, it follows that the div of a curl is zero for any $\textbf{F}$.
Now we show based on symmetry that $\operatorname{div}(\operatorname{curl}\textbf{F})=0$ for this $\textbf{F}$. Suppose we rotate our coordinate system by 180 degrees about the $y$ axis. This doesn't change the field or its components, but the curl lies in the $x$-$z$ plane, so the output of the curl operation has its components sign-flipped because of the new coordinate system used to describe it. This is an over-all reversal of the curl's direction, and since the div is linear, the effect is to sign-flip $\operatorname{div}(\operatorname{curl}\textbf{F})$.
But the divergence is a scalar, so the final result of taking $\operatorname{div}(\operatorname{curl}\textbf{F})$ cannot change just because we change our coordinates.
We have proved that $\operatorname{div}(\operatorname{curl}\textbf{F})$ flips its sign, but also that it doesn't change its sign. This is only possible if it is zero, as claimed.
