Prove that the $n$th difference of $x^n$ is $\sum _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n$ 
Prove that the $n$th difference of $x^n$ is $\displaystyle \sum \limits _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n$ (here the first difference is $\Delta (x^n):=(x)^n-(x-1)^n$ and the $n$th difference is defined recursively as $\Delta ^n(x^n):=\Delta ^{n-1}((x)^n)-\Delta ^{n-1}((x-1)^n)$).

I tried using induction, but just proving $\Delta ^n(x^n)$ is given by the indicated formula for all $n$ doesn't seem to work. Instead, I may need a more general formula involving $\Delta ^n(x^m)$, where $m$ and $n$ are positive integers. However, I'm not sure how to come up with this more general formula.
 A: This problem refers to the backward difference operator
\begin{align*}
\Delta\left(x^n\right):= x^n-(x-1)^{n}
\end{align*}
We can prove the formula
\begin{align*}
\color{blue}{\Delta^n\left(x^n\right)=\sum _{i=0}^n(-1)^i\binom{n}{i}(x-i)^n}\tag{1}
\end{align*}
using operator methods. We introduce the shift operator $E$ and identity operator $I$ with
\begin{align*}
Ef(x)&=f(x+1)\\
If(x)&=f(x)
\end{align*}
We can write the delta operator $\Delta$ as
\begin{align*}
\color{blue}{\Delta}\left(x^n\right)&=x^n-\left(x-1\right)^n\\
&=I\left(x^n\right)-E^{-1}\left(x^n\right)\\
&=\color{blue}{\left(I-E^{-1}\right)}\left(x^n\right)\tag{2}
\end{align*}
Since shift operator $E$ and identity operator $I$ are linear operators which commute we obtain according to (2) using the binomial theorem

\begin{align*}
\color{blue}{\Delta^n\left(x^n\right)}&=\left(I-E^{-1}\right)^n\left(x^n\right)\\
&=\sum_{i=0}^n\binom{n}{i}(-1)^iE^{-i}I^{n-i}\left(x^n\right)\\
&=\sum_{i=0}^n\binom{n}{i}(-1)^iE^{-i}\left(x^n\right)\\
&\,\,\color{blue}{=\sum_{i=0}^n\binom{n}{i}(-1)^i\left(x-i\right)^n}
\end{align*}
and the claim (1) follows.

A: Define $(Rf)(x) = f(x)-f(x-1)$. Note that $R$ is linear. It is straightforward using induction to show that
$(R^nf )(x) = \sum_{k=0}^n \binom{n}{k} (-1)^k f(x-k)$.
Abusing notation slightly, we see that for any constant $c$ we have $Rc = 0$ and that $Rx^p$ is a polynomial of degree $p-1$ with coefficient $p = \binom{p}{1}$. In particular,
$R^n x^n = n!$ and so
$ \sum_{k=0}^n \binom{n}{k} (-1)^k (x-k)^n = n!$.
