How can I prove that the set $\{1, x, x^2, ..., x^n, ...\}$ is a linearly independent set in $\mathbb{Q} [x]$ Were $\mathbb{Q}[x]$ is the set of all polynomial functions with coefficients in $\mathbb{Q}$.
I know that I have to prove that any finite combination of the set is linearly independent, but I do not how to approach this.
 A: Any finite subset of $\{1, x, x^2,\ldots\}$ is contain in $\{1, x, \ldots, x^n\}$ for some $n$. So it suffices to consider $\{1, \ldots, x^n\}$. Next, consider the Wronskian
\begin{align}
W[1,x, \ldots, x^n]=
\begin{vmatrix}
1 & x & x^2 & \ldots & x^n\\
0  & 1 & 2x & \ldots & nx^{n-1}\\
\vdots & 0 & 2 & \ldots & \vdots\\
\vdots & \vdots & 0 & \ldots &\vdots\\
0 & 0 & \ldots & 0 & n!
\end{vmatrix}=\prod^n_{k=0}k! \neq 0
\end{align}
which means $\{1, \ldots, x^n\}$ are linearly independent. 
A: If you view polynomials as formal expressions, then there's nothing to prove: any finite linear combination is of the form $c_1x^{n_1}+c_1x^{n_1}+c_2x^{n_2}+\cdots+c_kx^{n_k}$, where all $c_1,c_2,\ldots,c_k\in\mathbb{Q}$ and all $n_1,n_2,\ldots,n_k$ are distinct. By definition, the zero polynomial is the polynomial whose all coefficients are zero, end of story.
If you view polynomials as functions, then one possible proof can go as follows. Pick any finite subset, and let's say $x^n$ is the highest power in it. A linear combination of these monomials is a polynomial $c_nx^n+\text{(lower degree terms)}$. If we have such a linear combination that's identically equal to zero, then we have a polynomial of degree $\le n$ with more than $n$ roots (infinitely many roots — all rational numbers), which is impossible, unless this is the zero polynomial with all zero coefficients.
