Let $m_i\in \mathbb{N}, 1\leq i \leq n $ such that wlog if $m_i < m_j\in \mathbb{Z^+}$ then $i < j$. 
Consider the Vandermonde matrix $$V= 
\begin{bmatrix} 
1 & 1 & 1 & \ldots & 1 \\
x_1^{m_1} & x_2^{m_1} & x_3^{m_1} & \ldots & x_{n-1}^{m_1} \\
x_1^{m_2} & x_2^{m_2} & x_3^{m_2} & \ldots & x_{n-1}^{m_2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_1^{m_n} & x_2^{m_n} & x_3^{m_n} & \ldots & x_{n-1}^{m_n} \\
\end{bmatrix}$$
Under What conditions is the generalized Vandermonde Matrix Invertible? 
It is clear that if one lets $x_i=x_j$,then the determinant is Zero, and therefore the principle Vandermonde Determinant $V_p=\prod_{1 \leq i < j\leq n}^n(x_i-x_j)$ can be factored from the  $det(V)$. What remains is the schur function, which is a homogeneous polynomial of degree d which depends on $n$, which symmetric over its variables $x_i,1\leq i\leq n$. There is a proof of the fact that the coefficients of the schur function are positive integers which was given by Mitchel long ago. 

When one looks at the simplest case where $x_i \in \mathbb{C}, 1\leq i \leq n$ and $n=2$, 
$$V= 
\begin{bmatrix} 
1 & 1\\
x_1^n & x_2^n\\
\end{bmatrix}$$
one sees that the Vandermonde determinant is invertible when $\frac{x_1}{x_2}$ is not an $n^{th}$ root of unity. 

For $n=3$ things become difficult already as 
$$V_3= 
\begin{bmatrix} 
1 & 1 & 1\\
x_1^{m_1} & x_2^{m_1} & x_3^{m_1}\\
x_1^{m_2} & x_2^{m_2} & x_3^{m_2}
\end{bmatrix}$$
$det(V_3) = x_2^{m_2} x_1^{m_1}-x_3^{m_2} x_1^{m_1}-x_2^{m_1} x_1^{m_2}+x_3^{m_1} x_1^{m_2}-x_2^{m_2} x_3^{m_1}+x_2^{m_1} x_3^{m_2}$. 

As far as I see (which is not very far), characterizing the zero set of the this polynomial is not at all simple in either $\mathbb{C^3}$ or $\mathbb{R^3}$ with exception of the trivial case we already mentioned where $x_i=x_j$ for some $i\neq j$ and perhaps some roots of unity. One can guess that over $\mathbb{C^n}$, if the $|x_i|$ are distinct and have no-non trivial relations, then the schur function is non-zero. Which brings me back to my question:

What are the conditions under which the generalized Vandermonde matrix is invertible over $\mathbb{R^n}$ or $\mathbb{C^n}$? Is there Hope?

Well yeah, a little bit.There are some results one could see from Mitchel's result about the positivity of the coefficients of the Schur function. If we are working over $\mathbb{R^n}$, then as a consequence of the positivity of the coefficients mentioned by Mitchel, one can deduce that if $x_i$ are distinct and positive for $1 \leq i \leq n$, then the schur function takes on positive values and therefore non-zero. This result is mentioned in [this Paper][1]. 


Examples: 

1)Let $V_3$ be a the following $3$x$3$ matrix:
$$V_3= 
\begin{bmatrix} 
1 & 1 & 1\\
x_1 & x_2 & x_3\\
x_1^{3} & x_2^{3} & x_3^{3}
\end{bmatrix}$$
where $m_1 = 1, m_2 = 3$, then 
$det(V)=-(x_1-x_2) (x_1-x_3) (x_2-x_3) (x_1+x_2+x_3)$, so our schur function is the elementary symmetric function $x_1+x_2+x_3$ which is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

 2) 1)Let $V_3$ be a the following $3$x$3$ matrix:
$$V_3= 
\begin{bmatrix} 
1 & 1 & 1\\
x_1^{4} & x_2^{4} & x_3^{4}\\
x_1^{6} & x_2^{6} & x_3^{6}
\end{bmatrix}$$
where $m_1 = 4, m_2 = 6$, then 

$det(V)=-x_2^4 x_1^6+x_3^4 x_1^6+x_2^6 x_1^4-x_3^6 x_1^4+x_2^4 x_3^6-x_2^6 x_3^4$ 

$det(V)=-(x_1-x_2) (x_1+x_2) (x_1-x_3) (x_2-x_3) (x_1+x_3) (x_2+x_3) (x_1^2 x_2^2+x_3^2 x_2^2+x_1^2 x_3^2)$, so our schur function is $ (x_1+x_2)(x_1+x_3) (x_2+x_3) (x_1^2 x_2^2+x_3^2 x_2^2+x_1^2 x_3^2)$ which is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

 3) 1)Let $V_3$ be a the following $3$x$3$ matrix:
$$V_3= 
\begin{bmatrix} 
1 & 1 & 1\\
x_1^{4} & x_2^{4} & x_3^{4}\\
x_1^{7} & x_2^{7} & x_3^{7}
\end{bmatrix}$$
where $m_1 = 4, m_2 = 7$, then 

$det(V)=-(x_1-x_2) (x_1-x_3) (x_2-x_3)$ *

$(x_2^3 x_1^5+x_3^3 x_1^5+x_2 x_3^2 x_1^5+x_2^2 x_3 x_1^5+x_2^4 x_1^4+x_3^4 x_1^4+$
$2 x_2 x_3^3 x_1^4+2 x_2^2 x_3^2 x_1^4+2 x_2^3 x_3 x_1^4+x_2^5 x_1^3+x_3^5 x_1^3+$
$2 x_2 x_3^4 x_1^3+3 x_2^2 x_3^3 x_1^3+3 x_2^3 x_3^2 x_1^3+2 x_2^4 x_3 x_1^3+x_2 x_3^5$ $x_1^2+2 x_2^2 x_3^4 x_1^2+3 x_2^3 x_3^3 x_1^2+2 x_2^4 x_3^2 x_1^2+x_2^5 x_3 x_1^2+$
$x_2^2 x_3^5 x_1+2 x_2^3 x_3^4 x_1+2 x_2^4 x_3^3 x_1+x_2^5 x_3^2 x_1+x_2^3 x_3^5+$
$x_2^4 x_3^4+x_2^5 x_3^3)$ 

So our schur function is everything that we have after dividing out by $V_p$. This function is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

My goal is to really answer the following interesting question, Let $S\subset\mathbb{N}$ be infinite and arbitrary. Then Let the set $Z_d={zeros of the schur function with degree d|d\in S}$. Now Is $\Cup_d\in S Z_d$ empty, assuming that the variables of the schur function generate a free abelian group of rank equal to the number of variables. 

  [1]: https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=1xm-mGJ7IG9JPqcjmzTaxUd3yBUSmJqTFpH8tDaK4odyav5uwBqRiSk17eq4j&hl=en