Let $n_k$ be a sequence. We create a new sequence by taking the difference of consecutive terms in $n_k$. So the terms of the new sequence $a_k$ is defined as $a_i=n_{i+1}-n_i$. This is the difference operation on $n_k$. When we apply the difference operation to our new sequence j times, this is the jth order difference operation.
Proposition: Let $n^k$ be a sequence where $n$ is the index and $\Bbb N$ is the indexing set. Let $k\in \Bbb N$. If we apply the kth order difference procedure to the sequence, then we will have a sequence where all of the terms will be $k!$.
For an example, if we take the sequence $n^3$ and apply the 3rd order difference operation to it we get zeroth order: $\{1,8,27,64,125,...\}$ first order: $\{7,19,37,61,...\}$ second order $\{12,18,24,30,...\}$ third order: $\{6,6,6,...\}$.
I was looking at sequences of squares and cubes and I noticed that this more general statement was true. I want to know if someone has done this already and if they did who was it and where did they do it?
Yes, it is true. If you start with exponent $k,$ as in $n^k,$ the $k$th differences will be $k!$
As a result, if you begin with some polynomial $a_k n^k + a_{k-1} n^{k-1} + \cdots$ all the lower degree terms will have disappeared, and the final will be row $k$ with number $a_k k!$ That way you know what $a_k$ is, if you had enough terms to begin with. If you are sufficiently careful, you can subtract off $a_k n^k$ from the original sequence and proceed to find $a_{k-1},$ and so on.