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

Question

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

Angina Seng · Answer

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

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

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