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*}