Given an $n \times n$ vandermonde matrix $V$ which is invertible, is any $(n-1) \times (n-1)$ submatrix of $V$ invertible also? 
I think the answer is yes, but I don't know how to prove. 



Is any $(n-1)\times (n-1)$ submatrix of an $n \times n$ Vandermonde matrix invertible? 
This is an elaboration of Thierry's answer: If $0 \leq a\_0 < a\_1 < \cdots < a\_{d-1}$ is any sequence of nonnegative integers, then the determinant $\det \left( x\_j^{a\_i} \right)$ is equal to $\prod\_{i < j}(x\_i - x\_j) \cdot s\_{\lambda}(x\_1, \ldots, x\_d)$ where $\lambda = (a\_0, a\_1-1, a\_2-2, \ldots, a\_{d-1} - (d-1))$ and $s\_{\lambda}$ is the [Schur function](http://en.wikipedia.org/wiki/Schur_function). (Wikipedia takes this as the definition of Schur functions.) 


In your case, you want $(a\_0, a\_1, \ldots, a\_d)$ to be $(0,1,2,3,\ldots, d-k-1, d-k+1, d-k+2, \ldots, d)$. The corresponding partition $\lambda$ is $(0,0,0,\ldots, 1,1,\ldots, 1)$ where there are $k$ ones. This Schur function is the elementary symmetric function 
$$e\_k(x) := \sum\_{i\_1 < i\_2 < \cdots < i\_k} x\_{i\_1} x\_{i\_2} \cdots x\_{i\_k},$$
which is apparently also known as a Veite sum. 


So your minor vanishes precisely when there is a repeated column, or when $e\_k$ is $0$.


