Given a function $f(x)$ defined on $[a, a+nh]$ continuously differentiable up to $n$ times.

Let $P(x)$ be a polynomial of degree $n$ coincides with $f(x)$ on the $n+1$ points
$a, a+h, \ldots, a+nh$. It is clear

$$\left. \Delta_h^n ( f(x) - P(x) )\right|_{x=a}
=\sum_{k=0}^n (-1)^{n-k} \binom{n}{k}( f(a+kh)-P(a+kh) )
= 0
$$
This implies
$$\left.\Delta_h^n f(x)\right|_{x=a} = \left.\Delta_h^n P(x)\right|_{x=a} = h^n P^{(n)}(a)\tag{*1}$$

On the other hand, $f(x) - P(x)$ vanishes on $n+1$ points $a, a+h, a+2h, \ldots, a+nh$.

By Rolle's theorem, we can find $n$ points $x_1, x_2, \ldots, x_n$:

$$a < x_1 < a+h < x_2 < a+2h < \cdots a + (n-1)h < x_n < a+nh$$

such that $f'(x) - P'(x)$ vanishes on these $x_i$. Repeat applying Rolle's $(n-1)$ more times, we can conclude there is a $b \in (a, a+nh)$ such that

$$f^{(n)}(b) - P^{(n)}(b) = 0 \quad\iff\quad f^{(n)}(b) = P^{(n)}(b)\tag{*2}$$

Since $P(x)$ is a polynomial of degree $n$, $P^{(n)}(x)$ is a constant. This means $P^{(n)}(a) = P^{(n)}(b)$ and by combining $(*1)$ and $(*2)$, we get:

$$\frac{1}{h^n}\left.\Delta_h^n f(x)\right|_{x=a} = f^{(n)}(b)\quad\text{ for some }b \in (a, a + nh)$$