Why is $\{1,x,x^2,\ldots,x^n\}$ a basis for $P_n (F)$ where $P_n(F)$ is a set of polynomials with at most degree $n$ defined by Field axioms? Why is $\{1,x,x^2,\ldots,x^n\}$ a basis for $P_n (F)$ where $P_n(F)$ is a polynomial of degree $n$ defined by Field axioms?
I know how to show basis, where I need to show linear independence by proving every scalar, say $a_1=\cdots =a_n=0$ where $a$ $\in$ $F$. But how am I supposed to prove that? 
I was told that since scalars are unique, the only choice for, say $a_0x^0+ a_1 x^1+\cdots+a_n x^n=0$ is that scalars have to equal 0. Is it the way to do so?
 A: Please watch your language. It’s only a vector space that is said to have a basis. A single polynomial does not have a basis.
If, however, the set of all polynomials of degree $\le n$ is called $P_n(F)$, then that set is a vector space, and you may take as its basis the set you named, $\{1,x,x^2,\cdots,x^n\}$. This has nothing to do with whether $F$ is a finite or infinite field.
Now, you must make a distinction between a polynomial $f(x)\in P_n(F)$, which is nothing but an expression $\sum_0^na_jx^j$ on the one hand, and the associated function that takes an input $z$ and delivers the output $f(z)$ in whatever extension field $K\supset F$ that $z$ lies in, on the other hand. If your field is finite, then there are nonzero polynomials $f$ that will have an associated function that vanishes throughout $F$; but there will always be a larger field $K$ with an element $z\in K$ for which $f(z)\ne0$.
As a simple example, consider the polynomial $x^2+x$, with coefficients in the field $\Bbb F_2$ with only the two elements $0$ and $1$. This polynomial vanishes at both of these elements; but when you plug in an element of the field $\Bbb F_4$ with four elements that isn’t in $\Bbb F_2$, then the result is nonzero.
A: Let's take Somos' caution into account and assume that the field $F$ is infinite. Take a linear combination of $1, x, \ldots , x^n$ with at least one $a_i \neq 0.$ The result is a polynomial of degree $k$, where $k$ is the largest $i$ for which $a_i \neq 0.$ This polynomial can have at most $k$ roots in $F.$ Hence, it cannot equal the $0$ polynomial, which has infinitely many roots. Generally, this argument shows that no polynomial of non-negative degree at most $n$ can equal the $0$ polynomial. The only recourse is for $a_i=0$ for all $i.$ Hence, the elements $1, x, \ldots , x^n$ are independent.
This argument does not work for finite fields. In a finite field of $q$ elements, it turns out that all elements of the field satisfy $x^q-x=0.$
You have some detail to fill in, but this is the gist of an argument.
