Let $F$ be a field, and $\mathbb{K}$ be the field of fractions of the polynomial ring $R = F[x_{1},x_{2},\ldots x_{n}].$ 

Take $n$ mutually projections from $\mathbb{K}^{n}$ onto $1$-dimensional spaces, $\{E_{i}: 1 \leq i \leq n \}.$ If you like, let $E_{i}$ be projection onto the $1$-dimensional space of row vectors in $\mathbb{K}^{n}$ with $0$ in place $j$ for $j \neq i.$ Consider the linear transformation $X = \sum_{i=1}^{n} x_{i}E_{i}.$ 
I claim that $\{ I,X,X^{2},\ldots , X^{n-1} \}$ is linearly independent and has the same linear span as $\{E_{1},E_{2},\ldots, E_{n} \}.$ Clearly the former span is contained in the span of the idempotents. For each $i,$ let $P_{i} = \prod_{j \neq i} \frac{X-x_{j}I}{x_{i}-x_{j}}.$ Now $P_{i} \neq 0$ for any $i$ because $XE_{i} = x_{i}E_{i}$ for each $i.$ In fact, we have $P_{i}E_{j} = 0$ for $j \neq i,$ and $P_{i}E_{i} = E_{i}.$ Since $P_{i}$ is in the linear span of the $E_{k}s,$ we must have $P_{i} = E_{i}$ for each $i.$ Hence each $E_{i}$ is in the linear span of $\{I,X,\ldots, X^{n-1}\}.$ This implies that the rows of the Vandermonde matrix are linearly independent, and that its determinant divides $\prod_{i < j}(x_{i}- x_{j})^{2}.$ In fact, we see that  $E_{k} \prod_{i < j}(x_{i}-x_{j}) $ is an $R$-combination of $\{I,X,\ldots ,X^{n-1} \}$ for each $k,$ which implies that the determinant of the Van der Monde matrix divides $\prod_{i < j} (x_{i}-x_{j})$ in $R.$ Consideration of the coefficient of $X^{n-1}$ in each $E_{k}$ shows that this product  must also divide the determinant in $R.$