Proving $\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_j}\!\!,\;\;1\le j\!<\!n$ Let $p_n$ denote the $n$th prime number.
How could one prove that:
$$\prod \limits_ {k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_j}\!\!,\;\;1\le j\!<\!n$$

Examples:
$n=3812,\;j=81\qquad\implies\quad\large{\prod \limits _{k=81}^{3812} \,\frac{p_{k+1}}{p_k}= \frac{p_{3813}}{p_{81}} = \frac{35897}{419}}$
$n=20019,\;j=1\qquad\implies\quad\large{\prod \limits _{k=1}^{20019} \frac{p_{k+1}}{p_k} = \frac{p_{20020}}{p_{1}} = \frac{224993}{2}}$
$n=129181,\;j=35\quad\implies\quad\large{\prod \limits _{k=35}^{129181} \!\!\frac{p_{k+1}}{p_k}= \frac{p_{129182}}{p_{35}} = \frac{1715059}{149}}$

$$\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_n} \times \frac{p_{n}}{p_{n-1}} \dots \frac{p_{j+2}}{p_{j+1}}\times \frac{p_{j+1}}{p_j}=\frac{p_{n+1}}{p_j}$$
