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.
 A: The notion of rank you are looking for admits an interpretation as the tensor rank for form as in the comment of @John_Ma, but I think this is rather confusing. The example your are looking at is the case of a two--form, which, in each point, defines a skew symmetric bilinear form on the tangent space. Hence you can look at the (point-wise) rank of this bilinear form. In particular, "is of maximal rank" in this case just means "is non-degenerate in each point", i.e. if $\omega (\xi,\eta)=0$ for all $\eta$, then $\xi=0$. This is only possible in the case of even dimension, and if the dimesnion equals $2n$, then non-degeneracy is equivalent to the fact that the $n$-fold wedge product of $\omega$ with itself is non-zero and hence a volume form. 
A: 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  $2p$.
A: A differential 2-form $\omega$ on a manifold $M$ is a choice of a skew-symmetric bi-linear map $\omega_p:T_pM\times T_pM\to R$ on the tangent space at each point $p$ of the manifold. This gives a linear map $v_p\to \omega(v,.)$ from $T_pM\to T^*M$. Pick a basis of $T_pM$ and the corresponding dual basis for $T^*M$, the rank of the matrix of this map w.r.t your chosen basis is called the rank of the differential form at p. If the differential form is at least continuous, the rank is local constant and hence constant over connected component of M. Note that the non-degeneracy of $\omega$ implies the invertablity of the map $v_p\to \omega(v,.)$ from $T_pM\to T^*M$.. If the rank of the map $\omega_p:T_pM\times T_pM\to R$ is maximal, we say $\omega$ is of maximal rank.
