Proving $∆^nf(x_0;h_1,\cdots,h_n)=f^{(n)}(ξ)h_1\cdots h_n$

I had an idea that was longer than comment. But not a complete solution. I felt it may be helpful to you so I’ve provided it below:

This is essentially applying the mean value theorem with induction.

Let’s define the “divided difference” of a function as

$$D_h(f) = \frac{f(x+h)-f(x)}{h} $$

If we consider the interval $[x_0, x_0+h]$ (and the order here might be wrong since h could be negative, so we should really say “minimum interval containing both”) and let $f$ be differentiable over this interval then the mean value theorem applies and we have that there exists a $\eta \in [x_0, x_0+h]$ such that

$$ f’(\eta) = D_h(f)(x_0) = \frac{f(x_0 + h)-f(x_0)}{h} $$ Or in your notation (and you’ll see why we are doing this)

$$ h f’(\eta) = f(x_0 + h)-f(x_0) = \Delta f(x_0 : h)$$

You can go further and say that there is some function $\eta(x)$ such that:

$$ h f’(\eta(x)) = \Delta f(x:h)$$

And we have that $\eta(x)$ is always contained in the smallest interval containing $x$,$x+h$

So now the to continue building our intuition let’s Say we have some g(x) = $ \Delta f(x: h)$. It is the case that over the minimum interval containing $x_0, x_0+h, x_0+h_2$ that there is some $c$ in the interval such that

$$ \Delta(g:h_2)(x_0) = h_2 g’(c)$$

Filling in the definition of g we have that:

$$ \Delta(f: h_1 : h_2)(x_0) = h_2 ( \Delta f(x: h))’(c) $$

$$ \Delta(f: h_1 : h_2)(x_0) = h_2 h_1 (f’(\eta(x)))’(c) $$

Which means:

$$ \Delta(f: h_1 : h_2)(x_0) = h_2 h_1 f’’(\eta(c)) \eta’(c)$$

Now the problem is that $\eta’(c)$ at the very end. There should be some way to absorb it by showing that $f’’(\eta(x)) \eta’(x) = f’’(m) $ for some suitable choice of m, at which point you can then use induction on the strategy we have demonstrated to cover the arbitrary case.