Let $\alpha \in \operatorname{Alt}^{n-1}(\mathbb{R}^n)$. Show that there exists $\alpha_1, \dots, \alpha_{n-1} \in \operatorname{Alt}^1(\mathbb{R}^n)=(\mathbb{R}^n)^*$ such that $$\alpha = \alpha_1 \wedge \dots \wedge \alpha_{n-1}.$$
Here is my considerations. Let $\{e_1, \dots, e_n\}$ be the basis for $\mathbb{R}^n$ and $\{\varepsilon_1, \dots, \varepsilon_n\}$ the dual base for $(\mathbb{R}^n)^*$. Now $\operatorname{Alt}^{n-1}(\mathbb{R}^n)$ has a basis $$\{\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_{n-1}} \mid 1 \le i_1 < \dots < i_{n-1} \le n \}.$$
So $\alpha \in \operatorname{Alt}^{n-1}(\mathbb{R}^n)$ takes the form $$\alpha = \sum_{1 \le i_1 < \dots < i_{n-1} \le n} c_{i_1, \dots, i_{n-1}}\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_{n-1}}.$$
Can we figure out from here what these covectors should be?
It is certainly possible that your approach might work, but I'm unsure how. My idea is: let's prove that elements of form $\alpha_1 \wedge \alpha_2 \wedge \cdots \wedge \alpha_{n-1}$ form a vector subspace of $\operatorname{Alt}^{n-1}(\mathbb{R}^n)$. As the whole $\operatorname{Alt}^{n-1}(\mathbb{R}^n)$ has a basis consisting of such vectors (wedges of the standard basis are enough), we will obtain that the subspace is in fact the whole space.
Let $S$ be the set of all $\alpha_1 \wedge \alpha_2 \wedge \cdots \wedge \alpha_{n-1}$. It is easy to see that it is closed under scalar multiplication, so let's check what happens with addition. Take $\xi_1, \xi_2 \in S$. Then we can write $\xi_1$ as some $\alpha_1 \wedge \alpha_2 \wedge \cdots \wedge \alpha_{n-1}$; let $W_1$ be the span of all $\alpha_i$. We similarly define $W_2$. $W_1$ and $W_2$ are $(n-1)$-dimensional subspaces of $\mathbb{R}^n$, so there is a $(n-2)$-dimensional subspace $W$ lying inside $W_1 \cap W_2$ (in most cases equal to that intersection, but we do not want to consider the case $W_1=W_2$ separately).
Choose a basis $\beta_1, \beta_2,\ldots,\beta_{n-2}$ for $W$, and some elements $\gamma \in W_1 \setminus W$, $\delta \in W_2 \setminus W$. Then $\xi_1$ lies in one-dimensional space $\operatorname{Alt}^{n-1}(W_1)$, which is spanned by $\gamma \wedge \beta_1 \wedge \cdots \wedge \beta_{n-2}$. Hence we get
$$ \xi_1 = c_1 \gamma \wedge \beta_1 \wedge \cdots \wedge \beta_{n-2} $$
for some $c_1$, and similarly
$$ \xi_2 = c_2 \delta \wedge \beta_1 \wedge \cdots \wedge \beta_{n-2}. $$
Now
$$\xi_1+\xi_2 = (c_1 \gamma+c_2 \delta) \wedge \beta_1 \wedge \cdots \wedge \beta_{n-2}$$
and we are done.