How to get to this equality $\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1$?

Question

How to get to this equality $$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1?$$

I was studying the Euler Gamma function as it gave at the beginning of its history, and need to solve the following product operator, tried a few things, but I could not ...

$$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=\prod_{m=1}^{\infty} \frac{m^2+mx+m+x}{m^2+mx+m} =\prod_{m=1}^{\infty} \frac{m(m+x+1)+x}{m(m+x+1)}$$

WolframAlpha checked in and the result really is correct.

http://www.wolframalpha.com/input/?i=simplify+prod+m%3D1+to+inf+%5B%28m%2B1%29%2Fm+*+%28m%2Bx%29%2F%28m%2B1%2Bx%29%5D

Answer 1

$$\begin{align} \prod_{n=1}^N\frac{m+1}{m}\frac{m+x}{m+x+1}&=\left(\prod_{n=1}^N\frac{m+1}{m}\right)\left(\prod_{n=1}^N\frac{m+x}{m+x+1}\right)\\\\ &=\left(\frac{2}{1}\frac{3}{2}\frac{4}{3}\cdots \frac{N}{N-1}\frac{N+1}{N}\right)\left(\frac{1+x}{2+x}\frac{2+x}{3+x}\frac{3+x}{4+x}\cdots \frac{N-1+X}{N+x}\frac{N+x}{N+1+X}\right)\\\\ &=\frac{N+1}{1}\frac{1+x}{N+1+x}\\\\ &\to 1+x \,\,\text{as}\,\,N\to \infty \end{align}$$

Answer 2

If you write out the first few terms of the product, you'll see that most of the factors cancel out:

$$ \require{cancel} \begin{align} \prod_{m=1}^{\infty} \frac{m+1}{m}\cdot\frac{m+x}{m+x+1} &= \left(\frac{\cancel2}{1}\cdot\frac{x+1}{\cancel{x+2}}\right)\left(\frac{\cancel3}{\cancel2}\cdot\frac{\cancel{x+2}}{\cancel{x+3}}\right)\left(\frac{\cancel4}{\cancel3}\cdot\frac{\cancel{x+3}}{\cancel{x+4}}\right)\cdots\\ &=x+1 \end{align} $$