TITLE: A specific Vandermond matrix
QUESTION [5 upvotes]: Consider the Vandermond matrix 
$$
V (x_1, x_2, \ldots , x_n) =
\begin{pmatrix}
  1      & x_1    & x_1^2  & \cdots & x_1^{n-1} & x_1^n & x_1^{n+1} & \cdots \\
  1      & x_2    & x_2^2  & \cdots & x_2^{n-1} & x_2^n & x_2^{n+1} & \cdots \\
  1      & x_3    & x_3^2  & \cdots & x_3^{n-1} & x_3^n & x_3^{n+1} & \cdots \\
  \vdots & \vdots & \vdots & \ddots & \vdots    & \vdots & \vdots & \cdots \\
  1      & x_n    & x_n^2  & \cdots & x_n^{n-1} & x_n^n & x_n^{n+1} & \cdots
\end{pmatrix}
$$
with the specific choice of $x_1 = e^{\lambda_1}, \dots x_n = e^{\lambda_n}$, where $\lambda_1, \dots, \lambda_n$ are distinct real numbers.
I was wondering, if one will always get by picking arbitrary (not necessary consecutive) $n$ columns in the above "infinite" Vandermonde matrix linearly independent vectors.

REPLY [8 votes]: This works, as pointed out by Samuel in a comment. Here's an easy direct argument: We are claiming that no non-trivial polynomial
$$
p(x) = \sum_{j=1}^N a_j x^{n_j}
$$
with $N$ terms (let me call $N$ the pseudo-degree) has $N$ or more distinct positive zeros.
This is immediate from an induction on $N$. First of all, we can assume that $p$ starts with a constant term. Then, if $p$ had $N$ distinct positive zeros, its derivative would be a polynomial of pseudo-degree $N-1$ with at least $N-1$ positive zeros.