Show that $(\ln x)^n$ are linearly independent over polynomial.

Write $u=\ln x$ so $\sum_{k=0}^nu^kf_k(e^u)=0$. This is a linear combination of terms of the form $u^ke^{lu}$ with integers $k,\,l\ge0$. By the result you already have in exponentials, all terms vanish, so the $f_k$ do too, as required.

Another possible proof (which uses the same idea you explained in the question): let us suppose that exist some polynomials $f_0,f_1,f_2,\dots ,f_n$ not all being the $0$ poly such that $$\forall x\in (0;+\infty): \sum_{i=0}^{n}f_i(x)(\ln x)^i =0$$ We can consider without loss of generality the case in which at least one of the $f_i$ is not divisible by the $x$ poly (if not divide all the polynomials by $x^M$ where $M$ is the maximum integer such that $x^M \vert f_i$ for all $i$; the integer $M$ is well-defined because the $f_i$ are not all the $0$ poly). Now we will prove by induction that for all the $f_i$ polynomials we have $f_i(0)=0$ .

First step) $f_n (0)=0$: this is the consequence of $$\lim_{x\to 0^+}\frac{\sum_{i=0}^{n}f_i(x)(\ln x)^i}{(\ln x)^n}=0$$ by hypothesis and $$\lim_{x\to 0^+}\frac{\sum_{i=0}^{n}f_i(x)(\ln x)^i}{(\ln x)^n}=f_n(0)$$ by continuity.

Before proceeding with the inductive step, we will prove the following useful lemma: $$\lim_{x\to 0^+}x(\ln x)^{\aleph}=0$$ with $\aleph\in \mathbb{R}$. If $\aleph \in (-\infty;0]$ the proposition is trivial, so let us suppose $\aleph \in (0;+\infty)$. Thus we have

$$\lim_{x\to 0^+} x(\ln x)^{\aleph}=\lim_{y\to -\infty}e^y y^{\aleph}=\lim_{y\to -\infty} \frac{y^{\aleph}}{e^{-y}}=0$$

Inductive step) Let us suppose that for all $j\in (k;n]$ (with $k\in \mathbb{N}$) we have $f_j(0)=0$. This implies $$\lim_{x\to 0^+}\frac{\sum_{i=0}^{n}f_i(x)(\ln x)^i}{(\ln x)^k}=f_k(0)$$ by continuity, by inductive assumption and thanks to the lemma, and $$\lim_{x\to 0^+}\frac{\sum_{i=0}^{n}f_i(x)(\ln x)^i}{(\ln x)^k}=0$$ by hypothesis.

So for every $i\in [0;n]$ we have $f_i(0)=0$ contradicting the initial recruitment. Therefore the $f_i$ are identically $0$.