If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, then $n-4m$? 
If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, reduced fraction, then what is $n-4m$?

This is a question I dug up from an old JEE Main paper (India). It's very intriguing to me because, although it looks simple, I can't seem to find he sum in this question.
The numerator of the nth term of the series seems to be the product of the first n terms of odd numbers.
The denominator is similarly made but the sequence starts from 7.
I have no idea how to find the sum to n terms in this situation. If it were the sum of odd numbers and not the product, I could have done it easily.
Please explain to me how you find the product of n terms of a sequence 
and also the sum to n terms of the given sequence.
I have never been introduced to the product of n terms before, If you could give me a proper intuition for it, it would really make my christmas.

The following is not particularly clever but works.
Note that $$1\cdot 3 \cdot 5 \cdot \cdots \cdot (2N-1) = \frac{(2N)!}{2^N N!}$$
Using this (or just considering ratios of adjacent terms as suggested in comments), each term in the series is
$$\frac{120(2n)! (n+3)!}{n!(2n+6)!} = \frac{15}{(2n+1)(2n+3)(2n+5)} = \frac{15}{8}\left(\frac 1 {2n+1} - \frac 2 {2n+3} + \frac 1 {2n+5}  \right)$$
Using telescopic sums, one finds the sum of $N$ terms to be
$$\frac 1 4 -\frac{15}8 \left( - \frac 1 {2N+3} + \frac 1 {2N+5}\right)$$
and the answer can be calculated.

Edit: As Cameron pointed out in the comments, the telescopic nature of the sum is made clearer by realizing you are summing the terms $$\left(\frac 1 3 - \frac 1 5\right) - \left(\frac 1 5 - \frac 1 7\right),\quad \left(\frac 1 5 - \frac 1 7\right) - \left(\frac 1 7 - \frac 1 9\right),\quad \cdots$$
