Does this equality hold? How to prove it?

Question

$$ \prod_{k=1}^{n-1}(x/k+1) = n + \sum_{j=2}^{n}\binom{n}{j}\prod_{k=1}^{j-1}(x/k-1) $$ For any $n>1$, $0<x<1$. I have tested the case for $n=2,3,4$ and it holds. But cannot figure out a way to prove it. Tried induction, but it seems hard to get the result.

Answer 1

Both expressions are polynomials of degree $n-1$. Hence if they coincide for $n$ values of $x$, they are the same polynomial. Testing the non-negative integers less than $n$, on the left hand side we have

$$\prod_{k = 1}^{n-1}\biggl(\frac{r}{k} + 1\biggr) = \binom{n+r-1}{r}.$$

On the right hand side, for $r = 0$ we have

$$\sum_{j = 1}^{n} \binom{n}{j}(-1)^{j-1} = 1 - \sum_{j = 0}^n \binom{n}{j} (-1)^j = 1 - (1-1)^n = 1 = \binom{n-1}{0},$$

and for $1 \leqslant r < n$, we have

\begin{align} \sum_{j = 1}^{n-1} \binom{n}{j} \prod_{k = 1}^{j-1}\biggl(\frac{r}{k} - 1\biggr) &= \sum_{j = 1}^r \binom{n}{j}\binom{r-1}{r-j} \\ &= \sum_{m = 0}^{r-1} \binom{n}{r-m}\binom{r-1}{m} \\ &= \binom{n+r-1}{r} \end{align}

by Vandermonde's identity. So we have equality at the $n$ points $0,1,\dotsc,n-1$, and thus everywhere.

Answer 2

Firstly, $$ \prod_{k=1}^{j-1} \left( \frac{x}{k} - 1 \right) = \binom{x-1}{j-1} $$ (write out the factors to see this). Now, $$ \sum_{j=1}^n \binom{n}{j} \binom{x-1}{j-1} = \sum_{\ell=0}^{n-1} \binom{n}{n-\ell} \binom{x-1}{n-1-\ell} = \sum_{\ell=0}^{n-1} \binom{n}{\ell} \binom{x-1}{n-1-\ell}, $$ by changing the sum index around and using $\binom{n}{k} = \binom{n}{n-k}$. We can now apply the Chu–Vandermonde identity $$ \sum_{k=0}^N \binom{\alpha}{k}\binom{\beta}{N-k} = \binom{\alpha+\beta}{N} $$ to find $$ \sum_{j=1}^n \binom{n}{j} \prod_{k=1}^{j-1} \left( \frac{x}{k} - 1 \right) = \binom{n+x-1}{n-1} \\ = \left( 1+\frac{x}{n-1} \right) \dotsm \left( 1+\frac{x}{2} \right) \left( 1+\frac{x}{1} \right) = \prod_{k=1}^{n-1}\left( 1+\frac{x}{n-1} \right). $$ I think this is the simplest way to do it.