This is a few comments on David's excellent elaboration of Thierry's answer. 

We may as well assume that the minor eliminates the last row $[1\ x_n\  x_n^2\  \cdots  \ x_n^{n-1}]$.

- The expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter so it is easy to set it to zero and solve.

- A cute equivalent condition is that the values $x_i$ should be the $n-1$ distinct roots of a polynomial $$a_0x^{n-1}+a_1x^{n-2}+\cdots+a_{n-2}x+a_{n-1}$$ with $a_k=0.$

- It seems that  $e_0$ would be the zero function. But one should require $k>0$ as this corresponds to a minor which eliminates the last column. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial reformulation we see that there can't be enough distinct roots.  

I was glad to have this general result mentioned. It is interesting to see what be observed even without it.

- The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first column, we end up with an all zero row and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each row and what remains is again a Vandermonde matrix and nonsingular.

- If the minor eliminates neither the first nor the last column then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last columns are identical (all 1's). For the problem considered here,  this is the only way (up to scaling) to have a repeated column. Also, this is not a different case than $e_k=0$ since we just have the roots of $x^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$  Then it is especially fortunate to have the result so nicely described.