So trying to show the above and have tried using induction. 

If I let P(n) be the statement that "$\{\frac{1}{x}, \frac{1}{x^2}, \frac{1}{x^3}..., \frac{1}{x^n}$} is linearly independent". 

It can be shown that firstly, P(1) = $\frac{1}{x}$ is linearly independent because it is not the $0$ "vector". 

Next, I want to assume that P(k) is true, i.e. 

supposing that $\frac{a\_1}{x} + \frac{a\_2}{x^2} + \frac{a\_3}{x^3}... \frac{a\_k}{x^k} = 0 \implies a\_1 = a\_2 = a\_3 = ... = a\_k = 0$. 

Then suppose we have $\frac{a\_1}{x} + \frac{a\_2}{x^2} + ... + \frac{a\_k}{x^k} + \frac{a\_{k+1}}{x^{k+1}}$ and suppose it is 0, then I tried multiplying both sides by $x$ to get:

$a\_1 + \frac{a\_2}{x} + ... + \frac{a\_k}{x^{k-1}} + \frac{a\_{k+1}}{x^k} = 0$, but we know that the standard basis for polynomials is a linearly independent set. So $a\_{k+1}$ is 0. Then this works for all $n \in \mathbb{N}$ by induction. So it'll extend naturally into the basis with infinite cardinality. 

Is this how it is? Thanks!