Prove that $\left\{x^{\lambda_i}\right\}_{i=0}^n$ is a Chebyshev system on $(0,\infty)$ The definition of Chebyshev system is as follows.

A linearly independent system of $n$ basis functions $\varphi_1,\ldots,\varphi_n$ on $[a,b]$ is a Chebyshev system on $[a,b]$ if every non-trivial linear combination $\displaystyle\sum_{k=1}^n a_k\;\varphi_k$ has at most $n-1$ zeros on $[a,b]$.

Problem
The problem is to show that for any real $0=\lambda_0 < \lambda_1 < \ldots < \lambda_n$, the basis functions $\left\{x^{\lambda_i}\right\}_{i=0}^n$ form a Chebyshev system on $(0,\infty)$.
Approach
The first thing that I should prove is that $\left\{x^{\lambda_i}\right\}_{i=0}^n$ form a basis of the space $C(0,\infty)$, right? I don't know how to see that.
Assuming that they form a basis. I think that the proof of being a Chebyshev system is equivalent to prove that
$$\det\left(\begin{matrix}1 & x_1^{\lambda_1} & \ldots & x_1^{\lambda_n}\\ \vdots & \vdots &  & \vdots\\ 1 & x_{n+1}^{\lambda_1} & \ldots & x_{n+1}^{\lambda_n}\end{matrix}\right)\neq 0$$
for all pairwise distinct $x_1, \ldots, x_{n+1}\in(0,\infty)$ because it means that there is no linear combination that evaluated in one of those points is zero.
I need a bit of guidance in this exercise. Thank you
 A: As I mentioned in the comment, you do not need to show that $x^{\lambda}$ forms a basis of $C(0,\infty).$ Obviously, it is not going to be true since you cannot span $C(0,\infty)$ using finite number of functions. There is a beautiful theorem due to Muntz which sates that $\sum\frac{1}{\lambda_i}=\infty$ is a equivalent to the system  $span\{x^{\lambda_n}\}^{\infty}_{n=0}$ being dense in $C[a,b],$ $a>0.$
As to the original problem, just use Rolle's theorem and induction on $n.$ Indeed, base case is trivial. Now to prove the step, we assume that $f(x)=\sum_{i=0}^ka_ix^{\lambda_i}=0$ has $k+1$ zeros in $(0,\infty).$ By Rolle's theorem, $f'(x)=\sum_{i=0}^k(\lambda_i)a_ix^{\lambda_i-1}=x^{\lambda_1-1}(\sum_{i=1}^k(\lambda_i)a_ix^{\lambda_i-\lambda_1})$ has at least $k$ zeros in $(0,\infty).$ This contradicts to the fact that $\{x^{\lambda_i-\lambda_1}\}_{i=1}^{k}$ forms a Chebyshev system. 
A: If $0 < x_0 < x_1 < \ldots < x_n$ and $\lambda_0 < \lambda_1 < \ldots < \lambda_n$, then the determinant of the generalized Vandermonde matrix
$$
\left(
\begin{matrix}
x_0^{\lambda_0} & x_0^{\lambda_1} & \ldots & x_0^{\lambda_n}\\ 
x_1^{\lambda_0} & x_1^{\lambda_1} & \ldots & x_1^{\lambda_n}\\ 
\vdots & \vdots & \ddots & \vdots\\ 
x_{n}^{\lambda_0} & x_{n}^{\lambda_1} & \ldots & x_{n}^{\lambda_n}
\end{matrix}
\right)
$$
is strictly positive due to a special case of Theorem 1 published in the article [Shang-jun Yang, Hua-zhang Wu, Quan-bing Zhang, 2001].
Also, you may find interesting articles that are devoted to the construction of unique normalized Bernstein-like bases in extended Chebyshev Müntz spaces. For more details consider, e.g., the article [Marie-Laurence Mazure, 1999].
(Hopefully, you can download the cited papers.)
