Let $\alpha \in \Lambda^{p}L$, which is $p$-th power of $L$, where $L$ is linear space of dimension equal to $n$. Let us consider the following map $f_{\alpha} \colon L \rightarrow \Lambda^{p+1}L$ given with the formula $f_{\alpha}(\sigma)=\alpha \wedge \sigma$, where $\wedge$ is a wedge product.
Prove that if $\alpha, \beta \in \Lambda^{p}L$ and $p<n$, then: $$f_{\alpha} = f_{\beta} \iff \alpha = \beta.$$
The part where we assume $\alpha = \beta$ is easy, but what about another implication? Does anything come your minds? This was the first thing I thought about: $$\alpha_{1} \wedge \dots \wedge \alpha_{p} \wedge \sigma= \beta_{1} \wedge \dots \wedge \beta_{p} \wedge \sigma \iff (\alpha_{1} \wedge \dots \wedge \alpha_{p} - \beta_{1} \wedge \dots \wedge \beta_{p}) \wedge \sigma=0.$$I am stuck here, but maybe I don't see something very obvious about this.
Let $e_i$ be a basis for $L$, and assume $\alpha =\sum_I a_Ie_I$ where $I\subseteq \{1,2,\dots, n\}, \ |I|=p$ and $e_{\{i_1,\dots, i_p\}} =e_{i_1}\land\dots \land e_{i_p}$ with $i_1<\dots<i_p$.
We have to prove $f_{\alpha} =0 \implies \alpha=0$.
Assume some $a_I\ne 0$, then since $|I|=p<n$, there's an index $j\notin I$, and hence $a_I(e_I\land e_j)$ will be a nonzero term in $f_\alpha(e_j)$.