Let $\{\alpha^1,...,\alpha^p\}$ be a collection of linearly independent 1-forms. Show a 2-form $\beta$ is of the form \begin{align*} \beta=\gamma^1\wedge\alpha^1+\cdots+\gamma^p\wedge\alpha^p \end{align*} for 1-forms $\gamma^i$ if and only if \begin{align*} \beta\wedge\alpha^1\wedge\cdots\wedge\alpha^p=0. \end{align*}
I'm pretty new to wedge products. I really have not been able to come up with a strategy on how to connect the wedge product being zero with some sort of computation for the $\gamma^i$ with that said the forward direction of this proof I believe is straightforward.
If $\beta=\gamma^1\wedge\alpha^1+\cdots+\gamma^p\wedge\alpha^p$ then \begin{align*} \beta\wedge\alpha^1\wedge\cdots\wedge\alpha^p&=(\gamma^1\wedge\alpha^1+\cdots+\gamma^p\wedge\alpha^p)\wedge\alpha^1\cdots\wedge\alpha^p\\ &=\sum_{i=1}^p\gamma^i\wedge\alpha^i\wedge\alpha^1\wedge\cdots\wedge\alpha^p\\ &=\sum_{i=1}^p\gamma^i\wedge0\\ &=0. \end{align*}
Assuming this is correct I am very stuck on the reverse direction. I'm sure that I need to extend the linearly independent collection of 1-forms to a basis but beyond that I've just been spinning my wheels for quite some time. Any help would be greatly appreciated.
Extend your $\alpha_i$ to a basis in the dual space $\{\alpha_1,\alpha_2,\dots \alpha_p,\alpha_{p+1},\dots\alpha_n\}$. Then, your $2$-form $\beta$ can be written as $$ \beta=\sum\limits_{1\le i<j\le n}A_{ij}\alpha_i\wedge\alpha_j. $$ Now we can split the sum into two pieces, the first will contain all wedge products containing at least one of the $\alpha_i$ with $1\le i\le p$ and the second, all wedge products of the form $\alpha_k\wedge\alpha_r$ with $k,r>p$: $$ \beta={\rm{First\ \ piece}}+\sum\limits_{k,r}A_{k,r>p, k<r}\alpha_k\wedge \alpha_r $$ Next, right-wedge-multiply by $\alpha_1\wedge\alpha_2\dots\wedge\alpha_p$. Then the first piece disappears since at least two factors in the wedge products coincide, and you get $$ 0=\sum\limits_{k,r>p, k<r}A_{k,r}\alpha_k\wedge \alpha_r\wedge\alpha_1\dots\wedge\alpha_p. $$ On the right hand side, all the $p+2$-forms are linearly independent in the exterior $p+2$-power space and therefore $A_{k,r}=0$ for all $k,r>p$. This implies that $\beta$ is equal to the first piece, where all the addends contain some $\alpha_i$ with $i=1,2,\dots p$, $$ \beta =\sum\limits_{1\le i<j\le p}A_{ij}\alpha_i\wedge\alpha_j=\sum\limits_{j=2}^p\gamma_j\wedge\alpha_j $$ where $\gamma_j=\sum_i A_{ij}\alpha_i$.