Why does $\frac{n!n^x}{(x+1)_n}=\left(\frac{n}{n+1}\right)^x\prod_{j=1}^{n}\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^x$

Question

Why does $$\frac{n!n^x}{(x+1)_n}=\left(\frac{n}{n+1}\right)^x\prod_{j=1}^{n}\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^x$$ where the subscript n is the rising factorial in the left denominator

my attempt: the index n in the product indicates the indicates the term $\left(1+\frac{x}{j}\right)^{-1}$ may be some series of infinite geometric sums from the rising factorial but how? This is page 2 in Andrews, Askey, Roy Special Functions.

score 2 · Accepted Answer · answered Jul 23 '20 at 04:39

$$\prod_{j=1}^n\left(\frac{j+1}j\right)^x$$ telescopes and equals $(n+1)^x$. Multiplying by $$\left(\frac n{n+1}\right)^x$$ gives $n^x$. $$\prod_{j=1}^n\left(1+\frac xj\right)^{-1}=\frac{n!}{(x+1)(x+2)\cdots(x+n)} =\frac{n!}{(x+1)^{(n)}}$$ etc. (I prefer $x^{(n)}$ and $x_{(n)}$ for the rising and falling factorials resp.)

Why does $\frac{n!n^x}{(x+1)_n}=\left(\frac{n}{n+1}\right)^x\prod_{j=1}^{n}\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^x$

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