Prove that the $nth$ term of an arithmetic progression of order $k$ can be written in the form $an^k+bn^{k-1}+...+pn+q$. This question is from Courant Analysis Vol 1 End of Chapter Misc Question 8.  Earlier it is explained that arithmetic progression of order 1, is where the differences successive terms of the progression are constant.  For order 2, the differences of the differences are constant etc.  

Prove that the $nth$ term of an arithmetic progression of order $k$ can be written in the form $an^k+bn^{k-1}+...+pn+q$, where $a,b,...p,q$ are independent of n.

 A: I may have made some computational mistakes. However, I tried to show a way.
If you denote the arithmetic progression as
$$D_{1}^{0},D_{2}^{0},...,D_{n}^{0},...$$
the difference sequence can be denoted by
$$D_{1}^{1},D_{2}^{1},...,D_{n}^{1},...$$
the relations between the sequences are as below
1-
$$D_{n}^{1}=D_{n+1}^{0}-D_{n}^{0}$$
2- $$D_{n}^{0}=(\sum_{j=1}^{n-1}D_{j}^{1})+D_{1}^{0}$$
We assume that the arithmetic progression is of order $k$, which means that after $k$ times of creating difference sequences, we should have a sequence
$$D_{1}^{k},D_{2}^{k},...,D_{n}^{k},...$$
in which all the members $D_{n}^{k}$ are the same number $N$.
Now, using the relation 2, you can write the $n$-th number of the main arithmetic sequence as below.
$$D_{n}^{0}=\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1}\bigg((\sum_{i_k=1}^{i_{k-1}-1}D_{i_k}^{k})+D_{1}^{k-1}\bigg)$$
or
$$D_{n}^{0}=\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1}\bigg((\sum_{i_k=1}^{i_{k-1}-1}D_{i_k}^{k})\bigg)+\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1}\bigg(D_{1}^{k-1}\bigg)$$
or
$$D_{n}^{0}=\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1}\bigg((\sum_{i_k=1}^{i_{k-1}-1}N)\bigg)+\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1}\bigg(D_{1}^{k-1}\bigg)$$
or
$$D_{n}^{0}=N\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1} \sum_{i_k=1}^{i_{k-1}-1}1+D_{1}^{k-1} \sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1} 1$$
Finally, you need to know that such summations
$$\sum_{i_1=1}^{n-1}\sum_{i_2=1}^{i_1-1}...\sum_{i_{k-1}=1}^{i_{k-2}-1} 1$$
are only functions of $n$.
