Linear independence of standard basis vectors from Vandermonde style vectors Is it true a statement that
all $n$ dimensional vectors of the standard basis (e.g. $[1 \ 0 \ 0 \ ...]^T$,  $[0 \ 1 \ 0 \ ...]^T$ etc ..)  are linearly independent from the set of the $n-1$ vectors $v_k$ (dim. n)  of the form (I would  call them Vandermonde style  vectors):   
$[1  \ x_k \ x_k^2 \ ....\ x_k^{n-1}]^T$ 
where all $x_k$ are distinct non-zero numbers?
I.e. standard basis vectors don't belong to the subspace generated by these $(n-1)$  vectors  $v_k$..
Example for $n=3$, $x_1=2$, $x_2=3$:
$det{ 
\begin{bmatrix}
1 & 1 & 1 \\
0 & 2 & 3 \\
0 & 2^2 & 3^2 
\end{bmatrix}}$ = 6    
so the vector $i$ is linearly independent from the remaining column vectors. The vectors $j$, $k$ are independent too, what can be easily checked.
 A: The answer is yes for $e_1 = (1, 0, \ldots, 0)^T$. If you could write it as a combination $e_1 = \sum_{k=1}^{n-1} \lambda_k \, v_k$, the coefficients could not be all $0$. By using the projection onto the $(n-1)$ last factors (i.e. by forgetting the first coefficient), you would then obtain a nontrivial relation $\sum_{k=1}^{n-1} \lambda_k \widetilde v_k = 0$, where $\widetilde v_k = (x_k, x_k^2, \ldots, x_k^{n-1})$. This contradicts the fact that the $(\widetilde v_k)$ are a basis of $\mathbb C^{n-1}$, by a simple application of Vandermonde's determinant.
This argument also works for $e_n = (0,0,\ldots, 0, 1)^T$, but I still don't know the answer for $e_2, \ldots, e_{n-1}$.
For $2 \leq \ell \leq n-1$, you have corresponding vectors $v_k^{[\ell]} = (1, x_k, \ldots, x_k^{\ell-1}, x_k^{\ell+1}, \ldots, x_k^{n-1})^T$. For my argument to work, you have to answer the question: 

Do the $(v_k^{[\ell]})_{k=1}^{n-1}$ form a basis of $\mathbb C^{n-1}$?

In fact, this question is easily seen to be equivalent to yours. Note that it's also equivalent to wondering if a $(n-1)\times(n-1)$ minor of the Vandermonde determinant is always invertible. The answer to that must be known, but alas, not by me.

This last question has been asked on MO. The answer is not always yes: the matrix $\begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & x \\  1 & 1 & x^2\end{pmatrix}$ always has a $2\times 2$-minor with $1$'s everywhere, so it should answer your question.
In fact, the counterexample you get is quite simple: if $x_1 = 1$ and $x_2 = -1$, we have $\frac{v_1 - v_2}2 = (0, 1, 0)^T$.

Sorry for the "train of thought"-style answer, I really edited the answer at the same time I understood the problem.
