How to prove a $k$-$1$ differential form is simple I've been both trying to prove and looking for a proof in a couple of book and on the Internet, and I can't find it.
How can I prove that a $k$-$1$ differential form defined on a $k$ dimensional manifold is simple? That is, it can be written as a single wedge product of $k$-$1$ $1$-forms?
 A: Just using vector notation, we show that every $k-1$ form on a $k$ dimensional space is a wedge of $1$ forms. The key point is that the sum of two wedge forms is again a wedge form. (think the term is decomposable).
But this is trivial for if you have a the sum of two $k-1$ wedges,
$$x_1\wedge \cdots \wedge \hat{x}_i \wedge \cdots \wedge x_k
+x_1\wedge \cdots \wedge \hat{x}_j \wedge \cdots \wedge x_k
=\pm(x_i+x_j)\wedge x_1\wedge \cdots \hat{x}_i \cdots \hat{x}_i  \cdots \wedge x_k
$$ 
Now you can just write the same thing in the cotangent space of a manifold with functions as coefficients.
A: For any vector space $\mathbb{V}$ there is a canonical vector space isomorphism $\Phi: \Lambda^{n - 1} \mathbb{V}^* \to \mathbb{V} \otimes \Lambda^n \mathbb{V}^*$ whose inverse is the contraction map. So, since $\dim \Lambda^n \mathbb{V}^* = 1$, given $\alpha \in \Lambda^{n - 1} \mathbb{V}^*$, we can write $\Phi(\alpha)$ as a simple element $v \otimes \nu$. If we extend $v$ to a basis $(v, w_2, \ldots, w_n)$ of $\mathbb{V}$ such that $\nu(v, w_2, \ldots, w_n) = 1$, say, with dual basis $(\eta_1, \ldots, \eta_n)$, then by construction $$\alpha = \nu(v, \, \cdot \, , \cdots , \, \cdot \,) = \eta_2 \wedge \cdots \wedge \eta_n,$$ and in particular $\alpha$ is decomposable.
