Proof for this 'power triangle' Is there a proof for this triangle  Using factorial as a difference of powers? 
The first row is every consecutive integer raised to the power of n(5 here), but when you write the difference of them, in between any two numbers, continuously for n times(5 here), the nth value happens to be n!
$1    \quad   32  \quad   243   \quad  1024   \quad 3125   \quad 7776   \\ 
   \;\;\; 31   \quad  211 \quad    781  \quad   2101 \quad   4651 \\   
      \quad\quad  180   \quad  570    \quad 1320 \quad   2550    \\
           \quad\quad\quad\; 390  \quad   750   \quad  1230   \\ 
             \quad\quad\quad\quad\;   360  \quad   480\\
                   \quad\quad\quad\quad\quad\quad 120 $
    $ \quad  $  That is 
    $\\ \sum\limits_{k=0}^{n}(-1)^k\cdot\binom{n}{k}\cdot(n-k)^m=
 \begin{cases}
  0  & m<n\\
  n! & m=n\\
 \end{cases}$
P.s- Someone please try to help me edit this properly, I'm new to stackexchange
 A: With the help of exponential generating functions we show

the following is valid
  \begin{align*}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-k)^m
\begin{cases}
  0  & m<n\\
  n! & m=n\\
 \end{cases}
 \end{align*}
We obtain
  \begin{align*}
A(z)&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-k)^m\right)\frac{z^n}{n!}\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^m\right)\frac{z^n}{n!}\tag{1}\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}
\left.\frac{d}{dx^m}\left(e^{kx}\right)\right|_{x=0}\right)\frac{z^n}{n!}\tag{2}\\
&=\left.\frac{d}{dx^m}\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}
\left(e^{kx}\right)\right)\frac{z^n}{n!}\right|_{x=0}\\
&=\left.\frac{d}{dx^m}\left(\sum_{k=0}^{\infty}e^{kx}\frac{z^k}{k!}\right)
\left(\sum_{l=0}^{\infty}(-1)^l\frac{z^l}{l!}\right)\right|_{x=0}\tag{3}\\
&=\left.\frac{d}{dx^m}\left(e^{ze^x}\cdot e^{-z}\right)\right|_{x=0}\\
&=\left.\frac{d}{dx^m}e^{z\left(e^x-1\right)}\right|_{x=0}\\
&=z^m+a_{m-1}z^{m-1}+a_{m-2}z^{m-2}+\cdots\tag{4}\\
\end{align*}

Comment:


*

*In (1) we exchange $k \longleftrightarrow n-k$

*In (2) we represent $k$ as $\frac{d}{dx}e^kx$ evaluated at $x=0$

*In (3) we use the Cauchyproduct of formal exponential power series
\begin{align*}
\left(\sum_{k=0}^{\infty}a_k\frac{z^k}{k!}\right)\left(\sum_{l=0}^{\infty}b_l\frac{z^l}{l!}\right)=
\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\binom{n}{k}a_kb_{n-k}\right)\frac{z^n}{n!}
\end{align*}

*In (4) it is not hard to observe that derivating $e^{z\left(e^{x}-1\right)}$ at $x=0$ $m$ times gives a polynomial in $z$ with highest power $z^m$ and coefficient $1$.

We conclude according to (4) and using $[z^n]$ to denote the coefficient of $z^n$ that
  \begin{align*}
n![z^n]A(z)&=n![z^n]\left(z^m+a_{m-1}z^{m-1}+a_{m-2}z^{m-2}+\cdots\right)\\
&=\begin{cases}
  0  & m<n\\
  n! & m=n\\
 \end{cases}
\end{align*}
  and the claim follows.

