Why are these vectors linearly independent?

Question

For vectors $\begin{pmatrix} 1 \\\ 1 \\\ \vdots \\\ 1 \end{pmatrix}$, $\begin{pmatrix} x_1 \\\ x_2 \\\ \vdots \\\ x_n \end{pmatrix}$, $\begin{pmatrix} (x_1)^2 \\\ (x_2)^2 \\\ \vdots \\\ (x_n)^2 \end{pmatrix}$, $\ldots$, $\begin{pmatrix} (x_1)^m \\\ (x_2)^m \\\ \vdots \\\ (x_n)^m \end{pmatrix}$ for natural numbers $n$ and $m$ where $m <n$, my professor mentioned offhand that these were linearly independent when the $x_i$ were all distinct. I was wondering what a simple way to show this would be, and whether any 2 of the $x_i$ being the same implies linear dependence here.

So far, I have tried thinking of this intuitively, thinking that $x^m$ grows faster than a sum of smaller powers does; so if we find coefficients such that the entry of the vector corresponding to the $m$th power is equal, the coefficients applied to the smaller powers will be too large. However, I haven't gotten far with this and would appreciate help showing linear independence here.

The these vectors, when placed as columns in a matrix, form the first $m+1$ columns of a Vandermonde matrix. Vandermonde matrices are usually shown to be invertible using determinants. — Theo Bendit, Mar 17 '24 at 01:40
Assume $c_0\mathbf v_0 + c_1\mathbf v_1 + \cdots + c_m\mathbf v_m = \mathbf 0$ where $\mathbf v_i = \begin{pmatrix} (x_1)^i \\ (x_2)^i \\ \vdots \\ (x_n)^i \end{pmatrix}$, which means the polynomial $c_0 x^0 + c_1 x^1 + \cdots + c_m x^m$ has $n$ distinct roots ($n>m$), so the polynomial must be a constant zero. — peterwhy, Mar 17 '24 at 01:40
Adding a link to the invertibility of Vandermonde matrices (with equal numbers of rows and columns), and its current top answer proved that by the independence of all columns. This question just considers a subset of all columns. — peterwhy, Mar 17 '24 at 02:58

Why are these vectors linearly independent?

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