# Proof

• Let $n\ge 1$ be a natural number.
• By the Stirling numbers of the first kind and rising factorial powers \begin{align}x^\overline{n}&=\sum_{r=1}^n\left[\begin{array}{c}n\\r\end{array}\right] x^r\tag{1},\\x^\overline{n+1}&=\sum_{r=1}^{n+1}\left[\begin{array}{c}n+1\\r\end{array}\right] x^r\tag{2}.\end{align}
• Moreover, by definition of rising factorial powers \begin{align}x^\overline{q}(x+q)^\overline{p}&=x(x+1)(x+2)\cdots(x+q-1)\cdot\nonumber\\&\quad\cdot (x+q)(x+q+1)\cdots(x+q+(p-q)-1)\nonumber\\&=x(x+1)(x+2)\cdots(x+p-1)\nonumber\\&=x^\overline{p}\tag{3}\end{align} for any integers $p,q\in\mathbb Z.$
• For $p=n+1$ and $q=n,$ we get \begin{align}x^\overline{n+1}&=x^\overline{n}(x+n)^\overline{1}=xx^\overline{n}+nx^\overline{n}\tag{4}.\end{align}
• Result $(4)$ allows us to write $(2)$ in terms of $(1)$ as \begin{align}\underbrace{\sum_{r=1}^{n+1}\left[\begin{array}{c}n+1\\r\end{array}\right] x^r}_{=:A}&=\underbrace{x\sum_{r=1}^n\left[\begin{array}{c}n\\r\end{array}\right] x^r}_{=:B}+\underbrace{n\sum_{r=1}^n\left[\begin{array}{c}n\\r\end{array}\right] x^r}_{=:C}.\tag{5}\end{align}
• Now, \begin{align}B&=\sum_{r=1}^n\left[\begin{array}{c}n\\r\end{array}\right] x^{r+1}=\sum_{k=2}^{n+1}\left[\begin{array}{c}n\\k-1\end{array}\right] x^{k}=\sum_{k=2}^n\left[\begin{array}{c}n\\k-1\end{array}\right] x^{k} + \left[\begin{array}{c}n\\n\end{array}\right] x^{n+1},\tag{6}\\A&=\left[\begin{array}{c}n+1\\1\end{array}\right]x+\sum_{k=2}^n\left[\begin{array}{c}n+1\\k\end{array}\right] x^{k}+\left[\begin{array}{c}n+1\\n+1\end{array}\right] x^{n+1}\tag{7},\\C&=n\left[\begin{array}{c}n\\1\end{array}\right]x+\sum_{k=2}^nn\left[\begin{array}{c}n\\k\end{array}\right] x^{k}.\tag{8}\end{align}
• From $(5)$ it follows $0=A-C-B,$ and thus \begin{align}0&=\left(\left[\begin{array}{c}n+1\\1\end{array}\right]-n\left[\begin{array}{c}n\\1\end{array}\right]\right)x\nonumber\\&\quad+\sum_{k=2}^n\left(\left[\begin{array}{c}n+1\\k\end{array}\right]-n\left[\begin{array}{c}n\\k\end{array}\right]-\left[\begin{array}{c}n\\k-1\end{array}\right]\right)x^k\nonumber\\&\quad+\left(\left[\begin{array}{c}n+1\\k+1\end{array}\right]-\left[\begin{array}{c}n\\n\end{array}\right]\right)x^{n+1}.\tag{9}\end{align}
• Sufficient for $(9)$ are the conditions \begin{align}0&=\left[\begin{array}{c}n+1\\1\end{array}\right]-n\left[\begin{array}{c}n\\1\end{array}\right],\tag{10}\\0&=\left[\begin{array}{c}n+1\\k\end{array}\right]-n\left[\begin{array}{c}n\\k\end{array}\right]-\left[\begin{array}{c}n\\k-1\end{array}\right],\quad k=2,3,\ldots,n\tag{11}\\0&=\left[\begin{array}{c}n+1\\k+1\end{array}\right]-\left[\begin{array}{c}n\\n\end{array}\right]\tag{12}.\end{align}
• Given the conventions $\left[\begin{array}{c}n\\0\end{array}\right]:=0$ and $\left[\begin{array}{c}n\\k\end{array}\right]:=0$ for $k > n,$ we can re-write those as \begin{align}\left[\begin{array}{c}n+1\\1\end{array}\right]&=\left[\begin{array}{c}n\\0\end{array}\right]+n\left[\begin{array}{c}n\\1\end{array}\right],\tag{13}\\ \left[\begin{array}{c}n+1\\k\end{array}\right]&=\left[\begin{array}{c}n\\k-1\end{array}\right]+n\left[\begin{array}{c}n\\k\end{array}\right],\quad k=2,3,\ldots,n\tag{14}\\\left[\begin{array}{c}n+1\\k+1\end{array}\right]&=\left[\begin{array}{c}n\\n\end{array}\right]+n\cdot\left[\begin{array}{c}n\\n+1\end{array}\right].\tag{15} \end{align}
• $(13),$ $(14)$ and $(15)$ can be written as a single recurrence relation \begin{align} \left[\begin{array}{c}n+1\\r\end{array}\right]&=\left[\begin{array}{c}n\\r-1\end{array}\right]+n\cdot \left[\begin{array}{c}n\\r\end{array}\right],\nonumber\\ \left[\begin{array}{c}n\\n\end{array}\right]&:=1,\quad \text{ for }n\ge 1\nonumber\\ \left[\begin{array}{c}n\\r\end{array}\right]&:=0,\quad \text{ for }r=0 < n\text{ or }r > n.\tag{16} \end{align}

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### References

#### Bibliography

1. Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960