I know the coefficients in Taylor series are cooked up to be equal to the derivatives of the original function. But it's still a bit vague to me how information at one single point give the entire function... so I'm trying to derive it using MVT. Kindly have a look at the attached picture. 

[![processed image of computation scan](https://i.stack.imgur.com/zbtzb.png)](https://i.stack.imgur.com/zbtzb.png)
[original image](https://i.stack.imgur.com/xQuwR.jpg) 

From my attempt so far, I don't see any way of getting *factorial* terms in the denominators. Although it seems $\theta\_i = \frac{1}{i+1}$ should fix the problem, but this proof looks beyond my depth. Any help?

If I apply MVT again on $f''$ :
$$f(a+h) = f(a) + hf'(a) + h^2\theta\_1 f''(a) + h^3\theta\_1\theta\_2f'''(a+h\theta\_1\theta\_2\theta\_3) $$
for some $\theta\_i \in (0,1)$

My question is how to get rid of these $\theta\_i$...