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

If you write out the first few terms of the product, you'll see that most of the factors cancel out:
$$
\require{cancel}
\begin{align}
\prod_{m=1}^{\infty} \frac{m+1}{m}\cdot\frac{m+x}{m+x+1} &= \left(\frac{\cancel2}{1}\cdot\frac{x+1}{\cancel{x+2}}\right)\left(\frac{\cancel3}{\cancel2}\cdot\frac{\cancel{x+2}}{\cancel{x+3}}\right)\left(\frac{\cancel4}{\cancel3}\cdot\frac{\cancel{x+3}}{\cancel{x+4}}\right)\cdots\\
&=x+1
\end{align}
$$
