If $(1+x)^n =~~ ^nC_0 ~+~ ^nC_1~x ~+~ ^nC_2 x^2~+~\cdots+~ ^nC_n x^n $, then the value of :
$$\bigg(1+\dfrac {C_0}{C_1}\bigg) \bigg(1+\dfrac {C_1}{C_2} \bigg) \cdots \bigg(1+\dfrac {C_{n-1}}{C_n} \bigg)$$ is =?
Attempt:
I have been unsuccessful in most of my attempts with regard to this question.
Could someone please give me some hints on how to move forward.
Thank you for your help.
We have $\dfrac{\dbinom{n}{k-1}}{\dbinom{n}k} = \dfrac{n!}{(k-1)!(n-k+1)!} \cdot \dfrac{k!(n-k)!}{n!} = \dfrac{k}{n-k+1}$. Hence, we have $$1+\dfrac{\dbinom{n}{k-1}}{\dbinom{n}k} = 1 + \dfrac{k}{n-k+1} = \dfrac{n+1}{n-k+1}$$ Hence, we have $$\prod_{k=1}^n \left(1+\dfrac{\dbinom{n}{k-1}}{\dbinom{n}k}\right) = \prod_{k=1}^n \dfrac{n+1}{n+1-k} = \dfrac{(n+1)^n}{n!}$$
To supplement the first answer posted above, a more "visual" approach is given below.
Assume the following notation: $\displaystyle C_r= {^nC_r}=\binom nr$
From the identity $\displaystyle \binom nr+\binom n{r+1}=\binom {n+1}{r+1}$ we have $\displaystyle C_r+C_{r+1}= {^{n+1}C_{r+1}}$
Hence $$\begin{align} &\quad\bigg(1+\frac{C_0}{C_1}\bigg) \bigg(1+\frac{C_1}{C_2}\bigg) \bigg(1+\frac{C_2}{C_3}\bigg) \cdots \bigg(1+\frac{C_{n-1}}{C_{n}}\bigg)\\ &= \bigg(\frac{C_0+C_1}{C_1}\bigg) \bigg(\frac{C_1+C_2}{C_2}\bigg) \bigg(\frac{C_2+C_3}{C_3}\bigg) \cdots \bigg(\frac{C_{n-1}+C_n}{C_n}\bigg)\\ &= \frac{^{n+1}C_1}{\;\;^nC_1}\cdot \frac{^{n+1}C_2}{\;\;^nC_2}\cdot \frac{^{n+1}C_3}{\;\;^nC_3}\cdot \cdots \cdot \frac{^{n+1}C_n}{\;\;^nC_n} \\ &\color{lightgray}{= \frac{\frac{n+1}1} {\frac n1}\cdot \frac{\frac{(n+1)n}{2}}{\frac{n(n-1)}{1\cdot 2}}\cdot \frac{\frac{(n+1)n(n-1)}{1\cdot 2\cdot 3}}{\frac{n(n-1)(n-2)}{1\cdot 2\cdot 3}}\cdot\cdots\cdot \frac{\frac{(n+1)n(n-1)\cdots 2}{1\cdot 2\cdot 3\cdots n}}{\frac{n(n-1)(n-2)\cdots 1}{1\cdot 2\cdots n}}}\\ &= \frac{n+1}n\cdot \frac{n+1}{n-1}\cdot \frac{n+1}{n-2}\cdot\cdots\cdot \frac{n+1}{1}\\ &=\frac{(n+1)^n}{n!} \qquad\blacksquare \end{align}$$
