The original question is as follows:
Let $(f_1, f_2, …, f_k)$ and $(g_1, g_2, …, g_k)$ be linearly independent sets of dual vectors in $(\mathbb{R}^n)^*$.
Show that both sets span the same k-dimensional subspace of $(\mathbb{R}^n)^*$ iff
$f_1 \wedge f_2 \wedge\ …\ \wedge f_k = c \cdot g_1 \wedge g_2 \wedge\ …\ \wedge g_k$, $c \neq 0$.
My proof is as follows, but I am not sure that it is sufficient:
$(f_1, f_2, …, f_k)$ and $(g_1, g_2, …, g_k)$ are linearly independent in $(\mathbb{R}^n)^*$, so they form a basis for a k-dimensional subspace of $(\mathbb{R}^n)^*$ such that, for an ordered basis $(v_1, v_2, …, v_k)$ for a k-dimensional subspace of $\mathbb{R}^n$, $f_i(v_i) = g_i(v_i) = \delta_{ij}$ where $\delta$ is the Kronecker delta.
This is true iff $f_i(v_i)$ is some scalar multiple of $g_i(v_i)$, or $f_i(v_i) = c_j \cdot g_i(v_i)$, where $c_j \neq 0$.
From this we can deduce:
\begin{eqnarray*}
f_1 \wedge f_2 \wedge\ …\ \wedge f_k &=&
\begin{vmatrix}
f_1(v_1) & f_1(v_2) & … & f_1(v_k) \\
f_2(v_1) & f_2(v_2) & … & f_2(v_k) \\
\vdots & \vdots & \ddots & \vdots \\
f_k(v_1) & f_k(v_2) & … & f_k(v_k) \\
\end{vmatrix}
\\
&=&
\begin{vmatrix}
c_1 \cdot g_1(v_1) & c_1 \cdot g_1(v_2) & … & c_1 \cdot g_1(v_k) \\
c_2 \cdot g_2(v_1) & c_2 \cdot g_2(v_2) & … & c_2 \cdot g_2(v_k) \\
\vdots & \vdots & \ddots & \vdots \\
c_k \cdot g_k(v_1) & c_k \cdot g_k(v_2) & … & c_k \cdot g_k(v_k) \\
\end{vmatrix}
\\
&=&
c_1 c_2 … c_k
\begin{vmatrix}
g_1(v_1) & g_1(v_2) & … & g_1(v_k) \\
g_2(v_1) & g_2(v_2) & … & g_2(v_k) \\
\vdots & \vdots & \ddots & \vdots \\
g_k(v_1) & g_k(v_2) & … & g_k(v_k) \\
\end{vmatrix}
\\
&=&
c
\begin{vmatrix}
g_1(v_1) & g_1(v_2) & … & g_1(v_k) \\
g_2(v_1) & g_2(v_2) & … & g_2(v_k) \\
\vdots & \vdots & \ddots & \vdots \\
g_k(v_1) & g_k(v_2) & … & g_k(v_k) \\
\end{vmatrix}
\\
&=&
c \cdot g_1 \wedge g_2 \wedge\ …\ \wedge g_k
\end{eqnarray*}
Best Answer
I've since found a better solution to the problem:
Proof: $(\implies)$ Assume the two sets of vectors span the same $k$-dimensional subspace of $(\mathbb{R}^n)^*$. Then we may write each of the $f_i$ as a linear combination of the $g_j$ in a unique manner, since $\{g_1, g_2, ..., g_k\}$ is a basis for span$(g_1, g_2, ..., g_k)$. Write
\begin{eqnarray*} f_i = \sum_j c_{ij}g_j & & \textrm{for} \ 1 \leq j \leq k. \end{eqnarray*}
Observe that the $k \times k$ matrix $(c_{ij})$ is a change-of-basis matrix representing the basis $\beta_1 = \{f_1, f_2, \dots, f_k\}$ in terms of the second basis $\beta_2 = \{g_1, g_2, \dots, g_k\}$ for the space span$(g_1, g_2, \dots, g_k)$ = span$(f_1, f_2, \dots, f_k)$.
We now have
\begin{eqnarray*} f_1 \wedge f_2 \wedge \cdots \wedge f_k &=& \left(\sum_j c_{1j}g_j\right) \wedge \left(\sum_j c_{2j}g_j\right) \wedge \cdots \wedge \left(\sum_j c_{kj}g_j\right) \\ &=& \textrm{det}(c_{ij})g_1 \wedge g_2 \wedge \cdots \wedge g_k \end{eqnarray*}
$(\impliedby)$ Assume now that $f_1 \wedge f_2 \wedge \cdots \wedge f_k = c\ \cdot\ g_1 \wedge g_2 \wedge \cdots \wedge g_k$. By way of contradiction, let us assume additionally that span$(g_1, g_2, \dots, g_k) \neq$ span$(f_1, f_2, \dots, f_k)$. Then we may find a one-form $h \in$ span$(g_1, g_2, \dots, g_k) -$span$(f_1, f_2, \dots, f_k)$ such that $(f_1, f_2, \dots, f_k, h)$ is a linearly independent set, and so it follows that $f_1 \wedge f_2 \wedge \cdots \wedge f_k \wedge h \neq 0$. This leads to a contradiction:
\begin{eqnarray*} (f_1 \wedge f_2 \wedge \cdots \wedge f_k) \wedge h &=& c \cdot (g_1 \wedge g_2 \wedge \cdots \wedge g_k) \wedge h \\ &=& 0 \end{eqnarray*}
Because $\{g_1, g_2, \dots, g_k, h\}$ is a linearly dependent set. $\square$