I don't know how to approach this question from Flanders' Differential Forms. I see it was discussed here, but I don't believe that argument is correct and would apply to something like $2v_1 \wedge v_2+3v_1 \wedge v_3+6v_2 \wedge v_3$)
Let L be an n-dimensional space and $\alpha$ a 2-vector. Show that there is a basis $\sigma_1,\cdots,\sigma_n$ of L such that
$\alpha=\sigma_1\wedge\sigma_2 + \cdots + \sigma_{2r-1}\wedge \sigma_{2r}..$
Show that $\alpha^r\ne0$ and $\alpha^{r+1}=0$
I'm not sure if the book even defines what $\alpha^r$ means.
Assuming you are working with differential forms on $L$, the exterior algebra you would normally use is $\bigwedge(L^*)$, thought of as spaces of alternating multilinear maps on $V$. In particular, $\bigwedge^2(L^*)$ is viewed as the space of alternating bilinear maps on $L$. (Since you can view $\bigwedge^2(L)$ as alternating bilinear maps on $L^*$, however, this isn't an important distinction for the question you ask, but I find it easier to think about the bilinear maps!)
One way to show the result is to use induction on $\dim(L)$:
Given any $\alpha \in \bigwedge^2(L^*)$, then if $\alpha=0$, $r=0$ and there is nothing to prove. Otherwise, we may find $v_1,v_2\in L$ such that $\alpha(v_1,v_2)=-\alpha(v_2,v_1)\neq 0$. Rescaling $v_1$ as necessary, we may assume that $\alpha(v_1,v_2)=1$. Let $H_1 =\text{span}\{v_1,v_2\}$. Then if $H_1^{\perp} = \{v \in L: \alpha(v,h)=0, \forall h \in H_1\}$, and we take an arbitrary $v\in L$, then if we set
$$ h_v = \alpha(v,v_2)v_1 +\alpha(v_1,v)v_2 \in H_1, $$ we have $\alpha(v,v_2) = \alpha(h_v,v_2)$ and $\alpha(v_1,h_v) = \alpha(v_1,v)$, so that $v = h_v+ (v-h_v)$ and $v-h_v \in H_1^{\perp}$ and thus it follows that $L= H_1 \oplus H_1^{\perp}$.
Now on $H_1$, if $\delta_1,\delta_2$ denote the dual basis to $\{v_1,v_2\}$ then it is clear that $\alpha_{|H_1} = \delta_1\wedge \delta_2$. Moreover, by definition, $\alpha(v,w)=0$ if $v\in H_1, w \in H_1^{\perp}$, so that $\alpha$ is completely determined by its restrictions to $H_1$ and $H_1^{\perp}$.
But since $\dim(H_1^{\perp})\leq \dim(L)-2$, by induction we know that there is a basis $\{v_3,\ldots,v_n\}$ of $H_1^{\perp}$, such that $\alpha_{|H_1^{\perp}} = f_3\wedge f_4+ \ldots + \ldots f_{2r-1}\wedge f_{2r}$. But then it follows that $\alpha = \sum_{i=1}^r f_{2i-1}\wedge f_{2i}$ as required.
Finally, it is clear that $\alpha^r= r!f_1\wedge\ldots f_{2r} \neq 0$, and $\alpha^{r+1}=0$ since it is a $2r+2$-form, which evidently can only be nonvanishing on $V$, but as $\dim(V)=2r$, all alternating forms of degree greater than $2r$ vanish.