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Is there a fundamental proof of Taylor's theorem using little-o notation?

I assume $f:E\rightarrow F$ as a mapping between Banach spaces and write $(h^i)$ for $(h,\ldots,h)$ ($i$ times iterated).

The definition of the derivative gives $$f(x+h)=f(x)+f'(x)(h)+o(\|h\|).$$ Applying this to $f'$ I get $$f'(x+h)=f'(x)+f''(x)(h)+o(\|h\|)$$ where $o(\|h\|)$ has to be interpreted in the context of the corresponding ambient spaces, i.e. $E$ in the first line and $L(E,F)$ in the second line. Applying the second line to $h$ I get $$f'(x+h)(h)=f'(x)(h)+f''(x)(h^2)+o(\|h\|^2)$$ where the right hand looks at least similar to the right hand side of the equation I want to derive, namely $$f(x+h)=f(x)+f'(x)(h)+\frac{f''(x)(h^2)}{2}+o(\|h\|^2).$$ I could make it even more similar by writing $$f'(x+\frac{h}{2})(h)=f'(x)(h)+\frac{f''(x)(h^2)}{2}+o(\|h\|^2)$$ but I don't know how to proceed from there...

Edit: Sorry, what I wrote in my last (deleted) edit does not apply, I misread the text and the author addressed another formula with his quote!

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I found a proof in H Cartan's Differential Calculus. It doesn't exactly follow the route I attempted above, but it's still pretty fundamental, i.e. it essentially only uses the mean value theorem.Part 1Part 2

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