Here is my problem:
Given a vector x $\in \mathbb{R}^{n+1}$, the $(n + 1) \times (n + 1)$ matrix $V$ defined by
$v_{ij} = \begin{cases} 1 & \text{if } j = 1 \\ x_i^{j - 1} & \text{for } j = 2, \dots, n + 1 \end{cases}$
is called the Vandermonde matrix.
Suppose that $x_1, x_2, \dots, x_{n+1}$ are all distinct. Show that if c is a solution to $V$ x = 0, then the coefficients $c_1, c_2, \dots, c_{n+1}$ must all be zero, and hence $V$ must be nonsingular.
Each entry of $V$ c is an $n$th degree polynomial, so if $c_1, c_2, \dots, c_{n+1}$ were not all zero, then $x_1, x_2, \dots, x_{n+1}$ would have to be the roots of the polynomial. But it's perfectly possible for an $n$th degree polynomial to have $n$ roots, so I'm not sure how I can get this to work.