Representation of finite differences of order k We define recursively finite differences $ g_k (x) $ of order $ k $ of function $ f $ as follows:
$g_0(x)=f(x)$, $g_n(x)=g_{n-1}(x+h_n)-g_{n-1}(x) (n\in\mathbb{N})$.
It is known that all arguments of the function $ f $ (namely, $ x_0, x_0 + h_1, ... $) used in determining $ g_k (x_0) $ lie on the segment $ [a, b] \subset \mathbb {R} $ and the length of this segment is minimally possible, and also that $f \in C ^ {(n-1)} [a, b] $ and there is $ f ^ {(n)} (x) $ at least in the interval $ (a, b) $.
It is necessary to prove that then $\forall k\in \mathbb {N}$ there is a point $ \xi \in [a, b] $ such that $ g_k (x_0) = f ^ {(k)} (\xi) h_1 \cdot h_2 \cdot ... \cdot h_k $.
I suppose we should use Cauchy's mean value theorem, but do not have any idea how to do it
 A: This can be proved by induction on $k=0,1,2,\dots$. Indeed, let $g_0(f;x)=f(x)$, $g_n(f;x)=g_{n-1}(f;x+h_n)-g_{n-1}(f;x)$ for $n=1,2,\dots$. 
We want to show that for all $k=0,1\dots$ and all $x$ such that $a\le x\le x+h_1+\cdots+h_k\le b$ there is $c_k=c_k(f;x)$ such that $a<c_k\le c_k+h_1+\cdots+h_{k-1}<b$ and 
$$g_k(f;x) = f^{(k)}(c_k)h_1\cdots h_k. \tag{1}
$$
For $k=0$, this is trivial. Suppose this is true for $k=n$, where $n\in\{0,1,\dots\}$. Then, by the mean value theorem, for all $x$ such that $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ we have 
\begin{align}
g_{n+1}(f;x)&=g_n(f;x+h_{n+1})-g_n(f;x)\\
&=g_n'(f;b_n)h_{n+1}\\
&=g_n(f';b_n)h_{n+1} \\ 
&={f'}^{(n)}(c_{n+1})h_1\cdots h_nh_{n+1} \\ 
&=f^{(n+1)}(c_{n+1})h_1\cdots h_{n+1}   
\end{align}
for some $b_n=b_n(f;x)$ such that $x<b_n<x+h_{n+1}$ and $c_{n+1}:=c_n(f';b_n)$, so that $a<c_{n+1}\le c_{n+1}+h_1+\cdots+h_n<b$, as desired. 
The third equality in the above multi-line display is the crucial (even if very simple) observation, and the fourth equality in that display follows by induction, 
because the conditions $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ and $x<b_n<x+h_{n+1}$ imply 
$a\le x<b_n\le b_n+h_1+\cdots+h_n<x+h_1+\cdots+h_{n+1}\le b$, so that $a\le b_n\le b_n+h_1+\cdots+h_n\le b$. 
