Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ After seeing the neat little identity $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form
$\sum_{k=0}^a\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\epsilon_k=\pm1$. I discovered a relatively simple algorithm to generate these patterns: simply take $(n+2^p-1)^p$ and subtract $(n+2^p-2)^p$ (using $n\to n-1$) to get a polynomial of degree $p-1$. Take this difference and subtract $(n+2^p-3)^p-(n+2^p-4)^p$ (using $n\to n-2$) to get a polynomial of degree $p-2$. Repeat this process until $n^p$ is reach. The first few examples of this are
$$
\begin{align}
n^0&=1\\
(n+1)^1-n^1&=1\\
(n+3)^2-(n+2)^2-(n+1)^2+n^2&=4\\
(n+7)^3-(n+6)^3-(n+5)^3+(n+4)^3-(n+3)^3+(n+2)^3+(n+1)^3-n^3&=48
\end{align}
$$
Upon doing this for the next several powers and checking OEIS, it would appear the constant corresponding to the power $p$ is $$\large C_p=2^{\frac{p(p-1)}2} p!$$
However, this is an observation only, and I have no idea how to go about proving this. The only thing I notice is that $\frac{p(p-1)}{2}=\sum\limits_{k=1}^{p-1}k$, but I don't know how to use this fact. Does any one know how to prove this observation?

Note that since $R$ and $1$ commute,
$$
R^{2^k}-1=\left[\sum\limits_{j=0}^{2^k-1}R^j\right](R-1)\tag{1}
$$
Therefore,
$$
\begin{align}
\prod_{k=0}^{n-1}\left(R^{2^k}-1\right)x^n
&=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right](R-1)^nx^n\tag{2a}\\
&=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right]n!\tag{2b}\\
&=\left[\prod_{k=0}^{n-1}2^k\right]n!\tag{2c}\\[6pt]
&=2^{n(n-1)/2}\,n!\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $(1)$
$\text{(2b)}$: the $n^\text{th}$ forward difference of $x^n$ is $n!$
$\text{(2c)}$: on a constant sequence, $R^j=1$
$\text{(2d)}$: $\sum\limits_{k=0}^{n-1}k=n(n-1)/2$
