Proof for this 'power triangle'

Question

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

Answer 1

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.