Let $(x_1,\dots,x_k)$ be vectors in $\mathbb{R}^n$; let $X$ be the matrix $X = [x_1 \cdots x_k]$. If $I = (i_1, \dots, i_k)$ is an arbitrary $k$-tuple from the set $\{1,\dots, n\}$, show that $\phi_{i_1}\wedge\dots\wedge\phi_{i_k}(x_1,\dots,x_k) = \operatorname{det}X_I$ where $\wedge$ is the wedge product, and $\phi$ are the elementary $1$-tensors.
Here is what I know:
Given a basis $a_1,\dots,a_n$ for $V$, let $\phi_i$ denote the dual basis for $V^*$ and let $\alpha_i$ denote the corresponding elementary tensors. If $I = (i_1,\dots,i_n)$ is an ascending $k$-tuple of integers from the set $\{1,\dots,n\}$ then $\alpha_I = \phi_{i_1}\wedge\dots\wedge\phi_{i_k}$.
Let $\alpha_I$ be an elementary tensor on $\mathbb{R}^n$, where $I = (i_1,\dots, i_k)$. Given vectors $(x_1,\dots, x_k)$ of $\mathbb{R}^n$, let $X$ be the matrix $X = [x_1 \cdots x_k]$ then $\alpha_I(x_1,\dots,x_k) = \operatorname{det}X_I$ where $X_I$ denotes the matrix whose success rows are $i_1,\dots,i_k$
Proof Sketch: $\phi_{i_1}\wedge\dots\wedge\phi_{i_k}(x_1,\dots,x_k) = \operatorname{det}X_I$. What I'm thinking is that when we exchange rows the determinant changes sign that many times and similarly when we exchange $1$ tensors in a wedge product they get multiplied by $(-1)$ that many times, hence using $1$ and $2$ we can show the above required statement.
I think this is too simple and I'm missing something obvious, please provide feedback.