Summation equality with finite differences - prove with induction $f : \mathbb R^n \to \mathbb R, x,h \in \mathbb R^n$ and $m \in \mathbb N$, $m \geq 1$.
We wish to prove that $$\Delta_{mh}^r(f,x) = \sum_{k_1 = 0}^{m-1}\dots\sum_{k_r = 0}^{m-1}\Delta_h^r(f, x+k_1h+\dots + k_rh)$$
Where $\Delta_h^r (f,x):= \sum_{k = 0}^{r}(-1)^{r+k}\binom{r}{k}f(x+kh)$
We know this to be true when $r=1$, we wish to prove this with induction for all $r \in \mathbb N$. 
I've tried the natural thing to do but i got stuck, is there some trick here that I don't see?
 A: Everything becomes easy with the right notation and perspective. 
Let $L_h$ be the shift operator. Given a function $f:\mathbb R^n\to \mathbb R$, this means $L_hf$ is a new function $\mathbb R^n\to \mathbb R$, given by $(L_hf)(x)=f(x+h)$. This is related to $\Delta$ by
$$
\Delta^1_h(f,x) = f(x+h)-f(x)=(L_hf)(x)-f(x)
$$
Since we have equality for all inputs $x$, we have an equality of functions:
$$
\Delta^1_h f = L_hf - f=(L_h-1)f
$$
Furthermore, since holds for all functions $f$, this is an equality of operators:
$$
\Delta^1_h = L_h-1
$$
Here, $1$ is the identity operator, and the sum of operators is defined function-wise. Now, given any operator $G,$ let $G^n = G\circ G\circ \dots \circ G$ denote the composition of $G$ with itself $n$ times. Since operator arithmetic obeys all the usual laws, and composing two shift operators gives a different shift operator via $L_{h_1+h_2}=L_{h_1}\circ L_{h_2}$, we have
\begin{align*}
(\Delta^1_h)^r
    &= (L_h-1)^r\\
    &= \sum_{k=0}^r \binom{r}{k}L_h^{k}(-1)^{r-k}\\
    &= \sum_{k=0}^r \binom{r}{k}L_{hk}^{}(-1)^{r-k}\\
    &= \Delta_h^r
\end{align*}
This shows that $\Delta^r_h$ is the $r$-fold composition of $\Delta^1_h$ with itself, a good choice of notation!
Now, you already know that $\Delta_{mh}^1(f,x)=\sum_{k=0}^{m-1} \Delta_h^1(f,x+kh)$, as this is a telescoping sum. We can write this as an equality of operators:
$$
\Delta_{mh}^1=\sum_{k=0}^{m-1} \Delta^1_h\circ L_{kh}
$$
Now, take taking the $r$-fold composition of both sides, we get
\begin{align*}
    \Delta_{mh}^r
    &=\left(\sum_{k=0}^{m-1} \Delta^1_h\circ L_{kh}\right)^r
    \\&=\sum_{k_1=0}^{m-1}\sum_{k_2=0}^{m-1}\dots \sum_{k_r=0}^{m-1} \Delta^1_h\circ L_{k_1h}\circ  \Delta^1_h\circ L_{k_2h}\circ \dots \circ  \Delta^1_h\circ L_{k_rh}
\end{align*}
Finally, since $\Delta_h^1=L_h-1$, it is easy to show that $\Delta^1_h \circ L_{kh}=L_{kh}\circ \Delta^1_h$, i.e. that $\Delta^1_h$ commutes with all powers of $L_h$ (recall that $L_h^k=L_{kh}$). Therefore, we can rearrange this to
\begin{align}
\Delta_{mh}^r
&=\sum_{k_1=0}^{m-1}\sum_{k_2=0}^{m-1}\dots\sum_{k_r=0}^{m-1}
\Delta_h^1\circ\Delta^1_h\circ\dots \Delta^1_h\circ L_{k_1h}\circ L_{k_2 h}\circ \dots L_{k_rh} 
\\&=\sum_{k_1=0}^{m-1}\sum_{k_2=0}^{m-1}\dots\sum_{k_r=0}^{m-1}
\Delta_h^r\circ L_{k_1h}\circ L_{k_2 h}\circ \dots L_{k_rh} 
\end{align}
This equality of operators is exactly what you are trying to prove.
