let $K$ be a field (e.g., a finite field). Fix a dimension $n$, and, for $m > n$, $m$ distinct elements $v_1, \ldots , v_m$ of $K$. consider the following Vandermonde-like matrix:
\begin{equation} \begin{bmatrix} v_1 & v_1^2 & \cdots & v_1^n \\ \vdots & \vdots & \ddots & \vdots \\ v_m & v_m^2 & \cdots & v_m^n \end{bmatrix} \end{equation}
note that the powers, going rightward, are $(1, ..., n)$, not $(0, ..., n - 1)$ as in a usual Vandermonde matrix.
Finally, fix an element $b \in K$. For each $i \in \{1, \ldots , m\}$, write $H_i \subset K^n$ for the hyperplane $\{ (x_1, \ldots , x_n) \in K^n \mathbin{|} \sum_{j = 1}^n v_i^j \cdot x_j = b \}$.
Claim. If $b \neq 0$, then the hyperplanes $H_1, \ldots , H_m$ are in affine general position.
That is, each subset consisting of more than n of these hyperplanes has empty intersection. For a rigorous definition of affine general position, see [Stanley, p. 4].
This is false for $b = 0$, since these hyperplanes will then intersect in the origin.
Is the claim true? Why?
I am able to answer this in the affirmative using the following interesting weird trick. It's enough to assume that $m = n + 1$ (you can always winnow down the set $v_1, \ldots , v_m$ to your favorite $n + 1$-element subset, prove disjointness of the relevant hyperplanes using what I'm about to say, and then return back to the full set $H_1, \ldots , H_m$).
Thus let's rename the distinct elements to $v_0, \ldots , v_n$. If the intersection $\bigcap_{i = 0}^n H_i$ were nonempty—say, with $(y_1, \ldots , y_n) =: \mathbf{y} \in \bigcap_{i = 0}^n H_i$—then we'd have the following matrix equality:
\begin{equation} \begin{bmatrix} b \\ \vdots \\ b \end{bmatrix} = \begin{bmatrix} v_0 & v_0^2 & \cdots & v_0^n \\ \vdots & \vdots & \ddots & \vdots \\ v_n & v_n^2 & \cdots & v_n^n \end{bmatrix} \cdot \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix}. \end{equation}
expanding the domain of this linear mapping by one dimension and zero-padding the preimage, we deduce the following variant of this expression:
\begin{equation} \begin{bmatrix} b \\ b \\ \vdots \\ b \end{bmatrix} = \begin{bmatrix} 1 & v_0 & v_0^2 & \cdots & v_0^n \\ 1 & v_1 & v_1^2 & \cdots & v_1^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & v_n & v_n^2 & \cdots & v_n^n \end{bmatrix} \cdot \begin{bmatrix} 0 \\ y_1 \\ \vdots \\ y_n \end{bmatrix}. \end{equation}
indeed, for each fixed $b, v_0, \ldots , v_n$, the former and latter expressions are equivalent, over $\mathbf{y} \in K^n$.
now this matrix is a standard Vandermonde matrix, whose parameters (by assumption) are distinct. it's thus invertible, and its preimages are unique. yet because
\begin{equation} \begin{bmatrix} b \\ b \\ \vdots \\ b \end{bmatrix} = \begin{bmatrix} 1 & v_0 & v_0^2 & \cdots & v_0^n \\ 1 & v_1 & v_1^2 & \cdots & v_1^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & v_n & v_n^2 & \cdots & v_n^n \end{bmatrix} \cdot \begin{bmatrix} b \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{equation}
also holds, we thus conclude that the preimages $(b, 0, \ldots , 0) = (0, y_1, \ldots , y_n)$ are identical, an absurdity so long as $b \neq 0$. This contradiction completes the proof that $\bigcap_{i = 0}^n H_i = \varnothing$.