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$f : mathbb R^n to mathbb R, x,h in mathbb R^n$ and $m in mathbb N$, $m geq 1$.

We wish to prove that $$Delta_mh^r(f,x) = sum_k_1 = 0^m-1dotssum_k_r = 0^m-1Delta_h^r(f, x+k_1h+dots + k_rh)$$

Where $Delta_h^r (f,x):= sum_k = 0^r(-1)^r+kbinomrkf(x+kh)$

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?

Everything becomes easy with the right notation and perspective.

Let $L_h$ be the shift operator. Given a function $f:mathbb R^nto mathbb R$, this means $L_hf$ is a new function $mathbb R^nto mathbb R$, 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 $G^n = Gcirc Gcirc dots circ G$ 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 $L_h_1+h_2=L_h_1circ L_h_2$, we have
beginalign*
(Delta^1_h)^r
&= (L_h-1)^r\
&= sum_k=0^r binomrkL_h^k(-1)^r-k\
&= sum_k=0^r binomrkL_hk^(-1)^r-k\
&= Delta_h^r
endalign*
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:
$$
Delta_mh^1=sum_k=0^m-1 Delta^1_hcirc L_kh
$$
Now, take taking the $r$-fold composition of both sides, we get
beginalign*
Delta_mh^r
&=left(sum_k=0^m-1 Delta^1_hcirc L_khright)^r
\&=sum_k_1=0^m-1sum_k_2=0^m-1dots sum_k_r=0^m-1 Delta^1_hcirc L_k_1hcirc Delta^1_hcirc L_k_2hcirc dots circ Delta^1_hcirc L_k_rh
endalign*

Finally, since $Delta_h^1=L_h-1$, it is easy to show that $Delta^1_h circ L_kh=L_khcirc Delta^1_h$, 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
beginalign
Delta_mh^r
&=sum_k_1=0^m-1sum_k_2=0^m-1dotssum_k_r=0^m-1
Delta_h^1circDelta^1_hcircdots Delta^1_hcirc L_k_1hcirc L_k_2 hcirc dots L_k_rh
\&=sum_k_1=0^m-1sum_k_2=0^m-1dotssum_k_r=0^m-1
Delta_h^rcirc L_k_1hcirc L_k_2 hcirc dots L_k_rh
endalign
This equality of operators is exactly what you are trying to prove.