What is the rank of a differential form I've been searching the internet and books for a definition but none of the books on differential geometry and manifolds that I have contain the term rank in the index. 
While trying to find a definition I came across this (very bottom of page 394):
The differential form (2.7) is of maximal rank since by the criterion of nondegenracy there is a nonzero exterior power $\Omega^n = \pm k dp_1 \wedge \dots dp_n \wedge \dots $
 which up to the factor $\pm k$ is just the $2n$-volume form on $T^\ast M$.
It seems such a basic notion that I'm bewildered that I can't seem to find a definition. I'm very confused. I suspect that there is something wrong with the terms I search for but I don't know what else to search for. 

What is a differential form of maximal rank and what is the rank of a
  differential form?

I'm even confused about the most basic examples:
Consider for example differential one forms $\alpha, \beta, $. Then the differential form $\gamma = \alpha \wedge \beta$ has rank one because the minimal number of terms in the sum that constitutes $\gamma$ is just one term, namely, $\alpha \wedge \beta$. But the rank of $\gamma$ is two: it is a volume form in $\mathbb R^2$ and the maximal rank of any differential form in $\mathbb R^2$ is 2.

The definition of the rank of a skew-symmetric bilinear form $\beta$, can be found at page 111, in the book entitled "Tensor Analysis on Manifolds" by R.L. Bishop and S.I. Goldberg, Dover N.Y. 1980.
The rank of $\beta \in \bigwedge^2 V^\ast$, is the minimum number of the vectors in terms of which $\beta$ can be expressed. For example, if $\varepsilon^1,\cdots , \varepsilon^{2p}\in V^\ast$, are linearly independent vectors of the dual vector space, then the rank of 
$$ \beta = \varepsilon^1 \wedge \varepsilon^2 + \varepsilon^3 \wedge
\varepsilon^4 + \cdots + \varepsilon^{2p-1} \wedge \varepsilon^{2p}  \,,$$ 
is  $2p$.
