Prove Taylor expansion with mean value theorem

Equation (8) is certainly not true. A simple counterexample is $f(t)=t^3, x=0, \Delta=1,$ because we get $f'(\xi_1)=3\xi_1^2=f(1)-f(0)=1, \xi_1={1\over \sqrt3}$ and hence $\Delta_2=\xi_1={1\over \sqrt3}\neq {\Delta \over 2}.$ Also Taylor's Theorem doesn't say that an infinitely differentiable function necessarily coincides with its Taylor series. All Taylor's Theorem does is give a measure for the error made by estimating a function with its $n^{th}$ Taylor polynomial.

If you want to learn about the connection between the Taylor expansion and the Mean Value Theorem I suggest:
http://en.wikipedia.org/wiki/Taylor%27s_theorem
http://www.proofwiki.org/wiki/Taylor%27s_Theorem/One_Variable

A non-rigorous approach can be writing the power series and finding the coefficients through reasoning...

$$ f(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + \dots + a_n(x-a)^n + \dots $$

from there $a_0=f(a)$, $a_1= f'(a)$ and successive terms can be found by differentiating both sides further.

I'm not entirely sure what the exact proof is, but I would like to point something out. Let us take a look at:

$$\Delta_p = \frac{\Delta_1}{p}$$

I think on this one we have to think backwards. By using the mean value theorem we ended up with the intervals beyond $x$: $$x < x + \Delta_p + \Delta_{p-1} + \dots + \Delta_2 < x + \Delta_1$$ for all integers $p$. As each of these $\Delta_k$ is a individual limit; that is, we may choose how small the number beyond $x$ is for each $\Delta_k$(as long as it is follows the mean value theorem and it in fact stays in our bounded interval). Furthermore we may define: $$\Delta_k = \frac{\Delta_1}{p}$$ for $k\leq p$.

Now if it will work with other limits is beyond me, but I think this is a start to solving the problem. Perhaps if we used any other interval it would converge too fast and wouldn't be as accurate of an approximation? (I'm thinking in particularly of $\Delta_k = \frac{\Delta_1}{p^2}$ )