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.



A specific Vandermond matrix
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.


