Generating an independent set Suppose $f_1, f_2,...$ are a set of functions $\mathbb{R} \rightarrow \mathbb{R}$ so that each is the power of some non-constant function h. So $ f_ i=h^{n_i}$ for some natural number n.
Is it possible for such a set to be linearly dependent?
What additional conditions could ensure independence?
I was thinking about this because I need to show that a set in linearly independent. And I was not convinced that having the functions be powered lhjlj would be sufficient. 
 A: The following applies over an arbitrary field, with scalar valued functions on an arbitrary set (and was written before these were specified).  I will make the simplifying assumption that in fact $f_i = h^i$, which results in no loss of generality.  
The set $\{h,h^2,h^3,\ldots\}$ is linearly independent if and only if $h$ takes on infinitely many values.  (In particular, with the assumption that $h:\mathbb R\to\mathbb R$, it would suffice for $h$ to be nonconstant and continuous.)
First suppose that $h$ takes on infinitely many values, and let $n$ be a positive integer.  To see that $\{h,h^2,\ldots,h^n\}$ is linearly independent, suppose that $a_1,a_2,\ldots,a_n$ are scalars such that $a_1h+a_2h^2+\cdots a_nh^n=0$ identically.  Let $x_1,\ldots, x_n$ be points in the domain where $h$ takes distinct nonzero values, say $y_i=h(x_i)$, with $y_i\neq y_j$ if $i\neq j$.  Then the equations you get from plugging the $x_i$s into the identity $a_1h+a_2h^2+\cdots a_nh^n=0$ can be organized into the matrix equation
$$\begin{bmatrix}
y_1 & y_1^2 & \cdots & y_1^n\\
y_2 & y_2^2 & \cdots & y_2^n\\
\vdots & \vdots & \ddots & \vdots\\
y_n & y_n^2 & \cdots & y_n^n
\end{bmatrix}
\begin{bmatrix}
a_1 \\a_2\\ \vdots \\ a_n\end{bmatrix}=0.$$
If $A$ is the square matrix in this equation, note that $A$ has the same determinant as the invertible Vandermonde matrix 
$\begin{bmatrix} 1&\vec 0\\ \vec1& A\end{bmatrix}$, where $\vec{1}$ is the column vector of all $1$s and $\vec{0}$ is the row vector of all $0$s.  Thus, $A$ is invertible, which implies that $a_1=a_2=\cdots=a_n=0$.  
Now suppose that $h$ takes on only finitely many values, say $k$ of them, and that the sets where $h$ takes on distinct values are $E_1,E_2,\ldots,E_k$.  The vector space of functions that are constant on each $E_i$ is finite dimensional with dimension $k$, and each $h^i$ lies in this space.  Therefore $\{h,h^2,\ldots,h^{k+1}\}$ is linearly dependent.
