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I want to show that if $\sum_{k=0}^\infty a_kx^k = 0$ on $[0,1]$, then $a_k=0 \forall k\in\mathbb{N}$. I'm aware of the standard proof, but wanted to try another argument.

We know that a polynomial of degree $k$ has at most k roots, so the polynomial $\sum_{k=0}^N a_kx^k$ has at most $N$ roots. Now if we let $N\rightarrow \infty$ we know that we can have at most "$\aleph_0$ roots", but for the function to be identically zero we must have uncountably many roots.

I know that this argument is rubbish, but is there a way (by transfinite induction/Baire Category or something else) to make this idea rigorous?

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    Does this imply $f^{n}(0) = 0, \forall n \ge 1\implies a_k = 0, \forall k \ge 1$ ? – 2017-01-17
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    I don't understand what "this idea" you want to make rigorous is. – 2018-06-29
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    @AndrésE.Caicedo he's saying intuitively, a nonzero power series should have at most countably many roots, so if $f$ is zero on $[0,1]$, then all the coefficients should be zero. – 2018-06-29
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    @D_S But they want this as a corollary of the result about polynomials, so does the "idea" amount to "let $N \to\infty $", or is there actually something else? – 2018-06-29
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    You could use the [identity theorem](https://en.wikipedia.org/wiki/Identity_theorem), which is a bit like a generalization of this fact about polynomials. – 2018-06-29
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    @Jair isn't that the standard proof? The question wants a different approach. – 2018-06-29
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    What standard proof are you aware of? – 2018-06-29

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I think that the closest way to prove the claim following your idea is putting the problem in the complex plane, so you can use Rouché's theorem: basically, you approximate the behavior of $f$ in a small neighbourhood of $0$ by the first non-vanishing Taylor polynomial of $f$.

Suppose that for some $n\ge0$ we have that $a_n\neq0$. Get the smallest integer $n$ such that $a_n\neq0$. Then: $$\exists r>0,\forall |z|=r, |f(z)-a_n z^n|<|a_n|r^n.$$ Then, by Rouché's theorem, $f$ has exactly $n$ zeros in the disk $D_r$, contradicting the fact that $f$ is equal to zero in $[0,1]$.