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:
$n=3812,\;j=81\qquad\implies\quad\large{\prod \limits _{k=81}^{3812} \,\frac{p_{k+1}}{p_k}= \frac{p_{3813}}{p_{81}} = \frac{35897}{419}}$
$n=20019,\;j=1\qquad\implies\quad\large{\prod \limits _{k=1}^{20019} \frac{p_{k+1}}{p_k} = \frac{p_{20020}}{p_{1}} = \frac{224993}{2}}$
$n=129181,\;j=35\quad\implies\quad\large{\prod \limits _{k=35}^{129181} \!\!\frac{p_{k+1}}{p_k}= \frac{p_{129182}}{p_{35}} = \frac{1715059}{149}}$
 A: Expand out the product and cancel common factors:
$$\prod_{k=j}^n\frac{p_{k+1}}{p_k}=\frac{p_{j+1}p_{j+2}\cdots p_{n}p_{n+1}}{p_{j}p_{j+1}\cdots p_{n-1}p_{n}}=\frac{\color{red}{p_{j+1}}p_{j+2}\color{red}{\cdots p_{n}}p_{n+1}}{p_{j}\color{red}{p_{j+1}}\cdots p_{n-1}\color{red}{p_{n}}}=\frac{p_{n+1}}{p_j}$$
Note that this has nothing to do with $p_k$ being prime.
A: There's nothing mysterious going on here.  It's just a telescoping product.  You can prove these finite sums by induction.
The general case:
$$\prod_{i=j}^n \frac{a_{i+1}}{a_i} = \frac{a_{n+1}}{a_j}$$
Proof:
The proof will proceed by induction.  The base case $n=j$ is obvious.
Suppose our series does work for $n$.  Then $$\prod_{i=j}^{n+1} \frac{a_{i+1}}{a_i} = \prod_{i=j}^{n} \frac{a_{i+1}}{a_i} \cdot \frac{a_{n+2}}{a_{n+1}} = \frac{a_{n+1}}{a_j} \frac{a_{n+2}}{a_{n+1}} = \frac{a_{n+2}}{a_{j}}$$
A: $$\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_n} \times \frac{p_{n}}{p_{n-1}} \dots \frac{p_{j+2}}{p_{j+1}}\times \frac{p_{j+1}}{p_j}=\frac{p_{n+1}}{p_j}$$
