A curious identity from repeated differences of integer powers

Question

It is a well known fact that the repeated differences between $n$-powers of consecutive integers produce eventually $n!$. For example, for $n=3$ we have \begin{eqnarray} 1, 8 , 27, 64\\ 7, 19, 37\\ 12 , 18\\ 6 \end{eqnarray}

These repeated differences can be summarized in the identity $$ n!=\sum_{k=0}^n {{n}\choose{k}} (-1)^{n-k}(k+1)^n. $$

What is intriguing me is that the identity seems to be valid even when written as $$ n!=\sum_{k=0}^n {{n}\choose{k}} (-1)^{n-k}(k+c)^n, $$ where $c$ is any complex number!

I stumble upon this identity by pure chance and I verified it in many numerical experiments, but I could not prove it up to now. Probably I am missing some very basic fact and I would like if someone can tell me what it is.

Answer 1

You only need to put together three things:

1. That the $n^{th}$ repeated difference of $k^n$ equals $n!$
2. That the $(n+1)^{th}$ and subsequent repeated difference of $k^n$ all equal $0$.
3. That $(k+c)^n$ (thanks to old man Newton) is but a sum of $k^n$ and some lower powers of $k$ which all vanish to $0$ before we get to the $n^{th}$ repeated difference.