http://vapour-trail.blogspot.com/2006/03/brief-explanation-of-taylor-series-via.html provides an intuitive derivation of Taylor expansions from the mean value theorem that confuses me.

The derivation is described as follow. By the mean value theorem we have (assuming that $f$ has the differentiability properties requires for an infinite Taylor expansion)

$$f(x+\Delta)=f(x)+\Delta\cdot\frac{df}{dx}(\xi_1),\quad x<\xi_1<x+\Delta \tag4$$

Then one can reapply the mean value theorem to the first derivative in equation 4 to get

\begin{array} { l } \displaystyle { \frac { d f } { d x } \left( \xi _ { 1 } \right) = \frac { d f } { d x } ( x ) + \left( \xi _ { 1 } - x \right) \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) \quad ; x < \xi _ { 2 } < \xi _ { 1 } } \\ \displaystyle{ \Delta _ { 2 } \equiv \left( \xi _ { 1 } - x \right) } \\ \displaystyle{ \frac { d f } { d x } \left( \xi _ { 1 } \right) = \frac { d f } { d x } ( x ) + \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) } \\[2ex] \displaystyle f ( x + \Delta ) = f ( x ) + \Delta \cdot \frac { d f } { d x } \left( \xi _ { 1 } \right) \,\,\text{ from equation (4)}\\ \displaystyle { f ( x + \Delta ) = f ( x ) + \Delta \left[ \frac { d f } { d x } ( x ) + \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) \right] } \\ \displaystyle { f ( x + \Delta ) = f ( x ) + \Delta \cdot \frac { d f } { d x } ( x ) + \Delta \cdot \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) } \end{array}

Repeating the process indefinitely yields

$$ f ( x + \Delta ) = f ( x ) + \Delta \cdot \frac { d f } { d x } ( x ) + \Delta \cdot \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } ( x ) + \Delta \cdot \Delta _ { 2 } \cdot \Delta _ { 3 } \cdot \frac { d ^ { 3 } f } { d x ^ { 3 } } ( x ) + \ldots, \\ f(x+\Delta)=\sum _ { n = 0 } ^ { \infty } \left[ \left( \prod _ { p = 1 } ^ { n } \Delta _ { p } \right) \cdot \frac { d ^ { n } f } { d x ^ { n } } ( x ) \right], \quad \Delta_0=1 \,\,\&\,\, \Delta_1=\Delta $$

Now, as noted by the author, the last equation would be equivalent to an infinite Taylor series if we had

$$\Delta_p=\frac{\Delta_1}{p},\quad \forall p\in N\tag 8.$$

The author writes "Unfortunately, equation 8 is not as easy to derive/proof as its simplicity may otherwise suggest. As such, this article would end here." Someone asked for hints in the comments but the author never answered.

Could anyone provide a proof of (equation 8)?

It seems weird to me. For instance we have $\xi_1 = x + \Delta_2$. Equation 8 says $\Delta_2 = \frac{\Delta}{2}$. So $\xi_1 = x + \frac{\Delta}{2}$. Wouldn't that imply that for every $f$ (again having the required properties) and for every interval $[x,x+\Delta]$ in the domain of $f$, the mean value is always half way through the interval (i.e. the derivative at the half point in the interval is equal the the mean derivative over the interval)? This is obviously false (right?) so where is my mistake (if any)?