Electromagnetic Potential in Relativity Studying Special Relativity we discover that Maxwell's Equations can be also written in the following way:
$$\partial _\mu F^{\mu\nu}=\mu_0J^\nu$$
$$dF=0$$
Where: $F$ is the Electromagnetic Tensor, $J$ is the Four-Current Density and $dF$ to my understanding is simply a shortcut to write:
$$\partial _\lambda F_{\mu\nu}+\partial _\nu F_{\lambda\mu}+\partial _\mu F_{\nu\lambda}=0$$
Wonderful; but then my book goes on to say that since:
$$dF=0$$
We can state that:
$$F=dA$$
and so:
$$F=\partial _\mu A_\nu-\partial _\nu A_\mu$$
where $A$ is defined as the Electromagnetic Four-Potential. This last bit of reasoning makes no sense to me. Firstly I thought that $dF$ was just an abuse of notation, a shortcut; I cannot see how it can be interpreted as a differential. Secondly even if we take $dF$ as a proper differential why it should then imply that $F=dA$? And even then why on earth $dA$ should be equal to $\partial _\mu A_\nu-\partial _\nu A_\mu$?? I mean, just for starters, where did the minus sign came from?
 A: $dF$ is neither a shortcut nor an abuse of notation; it is the exterior derivative of the differential $2$-form $F$.
A $k$-form $G$ with vanishing exterior derivative $(dG=0)$ is called closed; a $k$-form which can be written as the exterior derivative of a $(k-1)$-form $(G = df)$ is called exact.
It is an elementary result in the theory of differential forms that all exact forms are closed.  That is, if we can write $G= df$, then we are guaranteed that $dG=0$.  However, the reverse is generically true only on star-shaped domains$^\dagger$.

Since the language of differential forms is new to you, note that you are used to dealing with similar concepts wrapped up in different notation.  The facts (i) that the curl of a gradient is always zero and (ii) that the divergence of a curl is always zero are specific examples of this idea.
Observing that $dF=0$ and concluding that $F=dA$ is spiritually the same as observing that $\nabla \times \mathbf E = 0$ and concluding that $\mathbf E = -\nabla \phi$, or observing that $\nabla \cdot \mathbf B = 0$ and concluding that $\mathbf B = \nabla \times \mathbf A$.

$^\dagger$As noted by Michael Seifert in the comments, this result can be extended more generally to contractible spaces.
