Show that $(\ln x)^n$ are linearly independent over polynomial. Question : If $f_n(x)(\ln x)^n + f_{n-1}(x)(\ln x)^{n-1} + ... +f_0(x) = 0$ where $ f_i (x) $are polynomial with real coefficients , then show that all $f_i (x)$ are identically zero.
I already know that if we replace $ \ln x $ as $ e^x$, then this is true.
Let $f_n(x)e^{nx}+ f_{n-1}(x)e^{(n-1)x} + ... +f_0(x) = 0$.
By divide $e^{nx}$, $$f_n(x)+ \frac{f_{n-1}(x)}{e^{x}}+ ... +\frac{f_0(x)}{e^{nx}}= 0$$.
Take limit infinity, $\lim f_n(x)=0$ and this mean $f_n(x)$ is identically zero.
By induction, we're done.
However, $(\ln x)$ is not accepted in this process because $\frac{P(x)}{\ln x}$ does not approach to zero as $ n \rightarrow \infty$ where $P(x)$ is polynomial. How to prove this?
 A: 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 $0$.
A: 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.
