Why is the set ${1,x^2,x^3...}$ linearly independent as functions? I know that the set is a basis for the vector space of polynomial over some fields. 
but why are they linearly independent?
namely, as function from the polynomial to the base field, what is a proof that they cannot be linearly dependent as functions?
 A: Polynomial functions $1,x,x^2,\ldots$ over a field $K$ are linearly independent over $K$ if and only if $K$ is infinite.  This is because $$x^{|K|}-x=0$$ for every $x\in K$, when $K$ is finite.  In fact, every function from $K$ to $K$ is a polynomial function when $K$ is finite, and in this case, the best you can say is that $1,x,x^2,\ldots,x^{|K|-1}$ are linearly independent functions.
On the other hand, abstract polynomials $1,x,x^2,\ldots$ over $K$ (i.e., they are elements of $K[x]$) are always linearly independent over $K$ (you can prove this claim, for example, via Vandermonde determinant).  You have to distinguish between polynomial functions and abstract polynomials.  Only when $K$ is infinite, you can roughly treat the two concepts as the same thing.
A: Two possible answers:


*

*The abstract definition of a polynomial ring stipulates that a polynomial, with coefficients in $\mathbf R$, say, is a sequence $(a_0,a_1,\dots, a_k,\dots)$ with finite support, i.e. there is only a finite number of $k$s such that $a_k\ne 0$. This set is endowed with an addition, defined componentwise, and a product defined as the Cauchy product. It makes the set of polynomials an  $\mathbf R$-algebra, and a basis as an $\mathbf R$-vector space is made up of the powers of the particular polynomial $(0,1,0,\dots, 0\dots)$, usually denoted as $X$.


If $n$ is the largest $k$ such that $a_k\ne 0$ (the degree of the polynomial),  the polynomial $(a_0,a_1,\dots,a_n,0,\dots, 0,\dots)$  can be rewritten as
$$a_0+a_1X+\dots+a_nX^n$$
so it is by construction that $a_0=a_1=\dots =a_n$.


*

*If there were a non-trivial  linear relation  between $1,x,\dots, x^n$:
$$a_0+a_1x+\dots a_n x^n=0,$$
the polynomial would have any value of $x$ as a root. However, this is impossible over an infinite field, as one shows  that a polynomial function $p(x)$ with coefficients in a field has  at most $\deg p$  roots.

A: You are having trouble proving this because it's not true. Over the $p$ element field  the polynomial $x^p -x$ is identically $0$ as a function, so $\{x, x^p\}$ is a dependent set.
Those polynomials are in fact dependent over any finite field, since 
there are only finitely many functions from a set to itself and the list of formal powers $x^n$ is infinite.
Note: here is an answer to the narrower question suggested in the comments. In characteristic $p$ the polynomials $1, x, x^2, \ldots, x^{p-1}$ are independent as functions. If some linear combination produced the $0$ function that linear combination would be a polynomial of degree at most $p-1$ with at least $p$ roots (since thare are at least $p$ elements in the field) hence the $0$ polynomial.
