Hodge star of second-rank antisymmetric tensor Say we have a tensor $F$ which just for familiarity's sake, we take to be a second rank antisymmetric tensor. I understand that given the Hodge star operator defined as 
$$^*F_{\alpha\beta}=\tfrac{1}{2}\varepsilon_{\alpha\beta\mu\nu}F^{\mu\nu},$$ applying it twice gives $^{**}F=-F.$ I'm not yet familiar enough with this notation to see intuitively why this should be true. Is there a quick way to understand and see how we can get this result? 
At a guess, I feel like this should perform a kind of "swapping" action thanks to the Levi-Civita tensor, because $^*F_{\alpha\beta}$ can only be nonzero when $\mu$ and $\nu$ do not repeat either $\alpha$ or $\beta,$ so we should have something like $^*F_{\alpha\beta}=F^{\mu\nu}.$ Then if $\varepsilon^{\alpha\beta\mu\nu}=-\varepsilon_{\alpha\beta\mu\nu},$ we might get $$^{**}F^{\alpha\beta}=\tfrac{1}{2}\varepsilon^{\alpha\beta\mu\nu}F^{\mu\nu}=-\tfrac{1}{2}\varepsilon_{\alpha\beta\mu\nu}F^{\mu\nu}=-F^{\mu\nu}.$$ Is it fine to then relabel $F^{\mu\nu}$ as $F^{\alpha\beta}$ and then conclude the result? I have a feeling this is a question with a very obvious answer, but I have not yet adjusted to this machinery.
Thanks for any help.
 A: Here's my approach to the question in a very low-machinery way.
First, we have
$$\star \star F_{\mu\nu} = \star \frac{1}{2}F^{\sigma\rho}\varepsilon_{\sigma\rho\mu\nu} = \frac{1}{4}F_{\alpha\beta}\varepsilon^{\alpha\beta\sigma\rho}\varepsilon_{\sigma\rho\mu\nu}$$
Because $\alpha\beta\sigma\rho$ and $\sigma\rho\alpha\beta$ have the same parity, we can write
$$\star \star F_{\mu\nu} = \frac{1}{4}F_{\alpha\beta}\varepsilon^{\sigma\rho\alpha\beta}\varepsilon_{\sigma\rho\mu\nu}$$
Now we want to consider the non-zero cases: given $\mu \neq \nu$, the only possibilities are
$$\alpha = \mu,\ \beta = \nu,\text{or}\ \alpha = \nu,\ \beta = \mu.$$
If $\sigma, \rho$ are determined and no two of $\sigma, \rho, \mu, \nu$ are equal, (no use of summation for the following line), we have
$$\varepsilon^{\sigma\rho\alpha\beta}\varepsilon_{\sigma\rho\alpha\beta} = -1,\ \text{and}\ \varepsilon^{\sigma\rho\alpha\beta}\varepsilon_{\sigma\rho\beta\alpha} = 1,\ \text{so}\ \varepsilon^{\sigma\rho\alpha\beta}\varepsilon_{\sigma\rho\mu\nu} = \delta^{\alpha}_{\nu}\delta^{\beta}_{\mu} - \delta^{\alpha}_{\mu}\delta^{\beta}_{\nu}$$
In reality, given $\mu\neq \nu$, we have exactly two choices for what $\sigma$ and $\rho$ can be (so that no two of the indices are equal). Thus (using summation again) our expression becomes
$$\frac{1}{4}F_{\alpha\beta}\varepsilon^{\sigma\rho\alpha\beta}\varepsilon_{\sigma\rho\mu\nu} = \frac{1}{4}F_{\alpha\beta}2(\delta^{\alpha}_{\nu}\delta^{\beta}_{\mu} - \delta^{\alpha}_{\mu}\delta^{\beta}_{\nu}).$$
But we may now evaluate this, and we have
$$\star\star F_{\mu\nu} = \frac{1}{4}F_{\alpha\beta}2(\delta^{\alpha}_{\nu}\delta^{\beta}_{\mu} - \delta^{\alpha}_{\mu}\delta^{\beta}_{\nu}) = \frac{1}{2}F_{\nu\mu} - \frac{1}{2}F_{\mu\nu} = - F_{\mu\nu}$$
