Tony Foster's Integer Powers in Pascal's Triangle

Problem

For integer $n,j\gt 1,$ we have

$\displaystyle \frac{\displaystyle j!\prod_{k=0}^{j}{n+k\choose k}}{\displaystyle \prod_{k=0}^{j-1}{n+1+k\choose k}}=(n+1)^j.$

Note that the multiplicands in the two products come from successive rows (or columns) of Pascals's triangle.

Solution 1

We start with an obvious $\displaystyle {n+1+k\choose k}=\frac{n+1+k}{n+1}{n+k\choose k}.$ Using that,

\displaystyle \begin{align} \frac{\displaystyle j!\prod_{k=0}^{j}{n+k\choose k}}{\displaystyle \prod_{k=0}^{j-1}{n+1+k\choose k}}&=\frac{\displaystyle j!\prod_{k=0}^{j-1}{n+k\choose k}\cdot {n+j\choose j}}{\displaystyle \prod_{k=0}^{j-1}\frac{n+1+k}{n+1}\cdot\prod_{k=0}^{j-1}{n+k\choose k}}\\ &=\frac{\displaystyle j!{n+j\choose j}}{\displaystyle \prod_{k=0}^{j-1}\frac{n+1+k}{n+1}}=\frac{(n+j)!}{n!}\cdot\frac{\displaystyle (n+1)^j}{\displaystyle\prod_{k=0}^{j-1}(n+1+k)}\\ &=\frac{(n+1)^j\cdot (n+j)!}{(n+j)!}=(n+1)^j. \end{align}

Solution 2

We'll use induction on $j.$ Let $\displaystyle F(j)=\frac{\displaystyle j!\prod_{k=0}^{j}{n+k\choose k}}{\displaystyle \prod_{k=0}^{j-1}{n+1+k\choose k}}$ and assume $F(m)=(n+1)^m,$ for some $m\gt 0.$ Then

\displaystyle \begin{align} \frac{F(m+1)}{F(m)}&=\frac{\displaystyle (m+1){n+m+1\choose m+1}}{\displaystyle {n+1+m\choose m}}\\ &= \frac{(m+1)(n+m+1)!}{n!(m+1)!}\cdot\frac{(n+1)!m!}{(n+1+m)!}=n+1, \end{align}

implying $F(m+1)=F(m)(n+1)=(n+1)^m(n+1)=(n+1)^{m+1}.$

Acknowledgment

This is Tony Foster's invention that he kindly shared on the CutTheKnotMath facebook page. Solution 1 is a slight modification of his original proof.

The above is a generalization of Tony's previous discovery.

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