Easiest ways to prove a list of polynomials is linearly independent What are some of the easiest, most methodical ways to prove the linear independence of $1,x,x^2,...,x^n$ in $\mathcal{P}(\textbf{F})$ for each nonnegative integer $n$.
I saw this method. Is this an appropriate way?
Assume that the list is linearly dependent.
Let $j$ be the largest index in $\{0,...,n\}$ such that $a_j \neq 0$
Then $a_0+a_1x+\dots+a_nx^n=0\implies$,$x^j=\frac{1}{a_j}(-a_0-\dots-a_{j-1}x^{j-1})$ expressing $x^j$ as a polynomial with degree $j-1$ a contradiction.
Other methods:
Since $a_0+a_1x+\dots+a_nx^n=0$ holds for all $n$,
this yields the system of equations 
$a_0=0$
$a_0+a_1+\dots+ a_n=0$
$a_0+2a_1+\dots+a_n(2)^n=0$
$\dots$
$a_0+a_1(n+1)+\dots+a_n(n+1)^n=0$
The determinant of the coefficient matrix of this system of equations is nonzero implying $a_1=\dots=a_n=0$
Is this method appropriate without using induction?
Also this method:
$a_0+a_1x+\dots+a_nx^n=0$
Plugging in $0$ yields $a_0=0$
differentiating both sides,
$2a_2+\dots+na_nx^{n-1}=0$
plugging in $0$ yields $a_n=0$ and so on.
Is this method appropriate without the use of induction?
If these methods need to be proved using induction(the last two) can someone show me how to do this. I am having trouble coming up with a way to prove the last two results by induction. Thanks
 A: There are two ways of defining polynomials over $\mathbb F$. 


*

*You define it as formal combinations $$\sum_{i=0}^n a_ix^i$$
and define de operations in the usual way.

*A polynomial is a function $f: \mathbb F \to \mathbb F$ given by $$f(x)=\sum_{i=0}^n a_ix^i.$$
The first definition is the one we use when working with general field. These definitions are not equivalent in general. For example, for $\mathbb F = \mathbb Z/2\mathbb Z$ we have the non trivial polynomial $f(x)=x^2+x$, but $f(0)=f(1)=0$. So, this $f$ is a non trivial polynomial, but it is the zero function if we think of $f$ as function.
Nevertheless, if $\mathbb F$ is infinite, then the two definitions coincide, as it is the case for $\mathbb F=\mathbb Q,\mathbb R,\mathbb C$, for example. In order to prove that we show the functions $x^i:\mathbb F \to \mathbb F$ are linearly independent. Consider a linear combination of the functions $x^i$: 
$$ f(x)=\sum_{i=0}^n a_i x^i = 0.
$$ 
We have to prove the coefficients $a_i$ are zero. Take $n+1$ pairwise distinct points $x_0, \ldots,x_n$ of $\mathbb F$, which exist because this field is infinite. Then we have the system
\begin{align*}
f(x_0)&=0 \\ 
f(x_1)&=0 \\
\vdots& \\  
f(x_n)&=0
\end{align*}
that can be written in matrix form as:
$$\begin{pmatrix} 1 & x_0 & x_0^2 & \cdots & x_0^n\\
1 & x_1 & x_1^2 & \cdots & x_1^n\\
\vdots & & \vdots & \ddots & \vdots\\
1 & x_n & x_n^2 & \cdots & x_n^n\\
\end{pmatrix}\begin{pmatrix} a_0\\
a_1\\
\vdots\\
a_n
\end{pmatrix}=0.
$$
The matrix $\begin{pmatrix} 1 & x_0 & x_0^2 & \cdots & x_0^n\\
1 & x_1 & x_1^2 & \cdots & x_1^n\\
\vdots & & \vdots & \ddots & \vdots\\
1 & x_n & x_n^2 & \cdots & x_n^n\\
\end{pmatrix}$
is called Vandermonde matrix and its determinant is 
$\Pi_{j<i}(x_i-x_j)$, which is non-zero, because the points $x_j$ are all pairwise distinct. Therefore, from the linear system above we conclude the $a_i$ are zero, and , therefore, the funcions $x^i$ are linearly independent.
