# Ascending Bases and Exponents in Pascal's Triangle

### Tony Foster 4 May, 2016

For integer $n\gt 1,\;$ let $\displaystyle P(n)=\prod_{k=0}^{n}{n\choose k}\;$ be the product of all the binomial coefficients in the $n\text{-th}\;$ row of the Pascal's triangle. Then

$\displaystyle\frac{\displaystyle (n+1)!P(n+1)}{P(n)}=(n+1)^{n+1}.$

To illustrate:

### Proof

$P(n)\;$ is given by

$\displaystyle P(n) = \prod_{k=0}^{n}{n\choose k}=\prod_{k=0}^n\frac{n!}{k!(n-k)!}=(n!)^{n+1}\prod_{k=0}^n\frac{1}{k!}\prod_{k=0}^n\frac{1}{(n-k)!}=(n!)^{n+1}\prod_{k=0}^n\frac{1}{(k!)^2}.$

Therefore,

\displaystyle\begin{align} \frac{(n+1)!P(n+1)}{P(n)} &= \frac{\left[(n+1)!\right]^{n+3}\displaystyle\prod_{k=0}^{n}(k!)^2}{\left[n!\right]^{n+1}\displaystyle\prod_{k=0}^{n+1}(k!)^2}\\ &=\frac{\left[(n+1)!\right]^{n+3}}{\left[n!\right]^{n+1}\left[(n+1)!\right]^2}\\ &=\frac{\left[(n+1)!\right]^{n+1}}{\left[n!\right]^{n+1}}\\ &=(n+1)^{n+1}. \end{align}

In other words, $\displaystyle\frac{P(n+1)}{P(n)}=\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^n}{n!}.$

(Note: some while later Tony cameup with a more general property of Pascal's triangle.)