Prove $\nabla\cdot(\nabla\times \vec{F})=\vec{0}$ using tensor notation.
Here is my shot at it:
$$\nabla\cdot(\nabla\times \vec{F})=\vec{0}$$ becomes $$\partial_{i}(\epsilon_{ijk}\partial_{j}F_{k})$$
Using the product rule.
$$\epsilon_{ijk}[F_{k}(\partial_{i}\partial_{j})+\partial_{j}(\partial_{i}F_{k})] = \epsilon_{ijk}F_{k}(\partial_{i}\partial_{j})+\epsilon_{ijk}\partial_{j}(\partial_{i}F_{k})$$
After permutation.
$$\epsilon_{jki}F_{i}\partial_{j}\partial_{k}-\epsilon_{kji}\partial_{j}\partial_{k}F_{i}$$
So wouldn't this look like $\vec{F}\cdot(\nabla\times \nabla)-\vec{F}\cdot(\nabla\times \nabla)=\vec{0}$? I am pretty sure you are not allowed to cross the gradient operator with itself. I don't think this is right.
Best Answer
You've rewritten $\partial_i(\partial_jF_k)$ as $F_k\partial_i\partial_j+\partial_j\partial_iF_k$. That would work if $\partial_j$ were an ordinary quantity you just multiply by $F_k$, but of course it's not. Indeed, your strategy also requires acknowledging $\partial_i$ is instead a differential operator, obeying the famous product rule.
The correct treatment needs no product rule. As @DavideMorgante's answer noted, you can just use the same symmetric indices argument in the proof of $A\cdot A\times F=0$ for a "normal" (i.e. non-operator-valued) vector $A$, since $\partial_i\partial_j=\partial_j\partial_i$ is just as true as $A_iA_j=A_jA_i$.