Prove that $ X_a=\{x_a; x_a=(a,a^2,a^3...), a \in R \}$ are linearly independent. I need to prove that all vectors in the set $X_a=\{x_a; x_a=(a,a^2,a^3...), a \in \mathbb{R} \} \subset \mathbb{R}^{\infty} $are linearly independent
I have tryed to prove it by induction but I don't know how to formulate it. I have tryed also to prove it by using polynomials (I know how to show that they   are linearly independent) but these are just scalars of a specific form not $(x,x^2,x^3...)$ where $x$ is a variable (and then so and so hold for every $x$) 
any Ideas? 
 A: I assume you mean $a\ne 0$, otherwise it is of course not true. Well, let $a_1,a_2,...,a_n$ be different non zero real numbers and write a linear combination $\lambda_1x_{a_1}+\lambda_2x_{a_2}+...+\lambda_nx_{a_n}=0$. Here the zero vector is $x_0$, all its coordinates are zeros, including the first $n$ coordinates of course. So we get a system of linear equations:
$\lambda_1a_1+\lambda_2a_2+...+\lambda_na_n=0$
$\lambda_1a_1^2+\lambda_2a_2^2+...+\lambda_na_n^2=0$
.................................................
$\lambda_1a_1^n+\lambda_2a_n^2+...+\lambda_na_n^n=0$
This is a system of linear equations with $\lambda_1,...,\lambda_n$ being the variables. To show $\lambda_1=...=\lambda_n=0$ it is enough to prove that the determinant of the matrix which represents the system is not zero. Write the matrix and see that its determinant is $a_1a_2...a_n$ times the determinant of a Vandermonde matrix which I assume you know is not zero. (of course we are using the fact that $a_1,a_2,...,a_n$ are all different) 
A: Assume that there exist a nonempty minimally sized set of nonzero elements 
$$\mathcal{S} = \left\{(x_0, x_0^2, x_0^3\ldots), \ldots , (x_n, x_n^2, x_n^3 \ldots)\right\}
$$
that are linearly dependent. Then
$$x_0^k = \sum_{i=1}^{n} c_i x_i^k$$
for all $k \in \mathbb{N}_0$. Then
$$\sum_{i=1}^{n} (c_ix_i) x_i^{k-1} = x_0^k = x_0(x_0)^{k-1} = \sum_{i=1}^{n} (x_0c_i) x_i^{k-1},$$
which implies that for all $k \in \mathbb{N}$,
$$\sum_{i=1}^{n} (x_0c_i - x_ic_i)x_i^{k-1} = 0$$
Since $\mathcal{S} \setminus \{(x_0, x_0^2, x_0^3, \ldots )\}$ is by assumption linearly independent, it must be that for each $i$, $x_0c_i - x_ic_i = 0$, thus $x_0 = x_i$, or $c_i = 0$. In other words, either
$$x_0 = x_1 = \ldots = x_n,$$
or $\mathbf{0} \in \mathcal{S}$. 
Contradiction.
