There's a quick trick!

In general, if we have the following:

$S=\left(1+\frac{1}{m}\right)\left(1+\frac{1}{m+1}\right)\cdots \left(1+\frac{1}{n-1}\right)\left(1+\frac{1}{n}\right)$

$S$ can represented as the following product:

$S=\prod_{i=m}^n \left(1+\frac{1}{i}\right)=\prod_{i=m}^n \left(\frac{i+1}{i}\right)\\ \implies S=\frac{\displaystyle\prod_{i=m+1}^{n+1} i}{\displaystyle\prod_{i=m}^n i}=\frac{(n+1)\cdot \displaystyle\prod_{i=m+1}^ni}{m\cdot \displaystyle\prod_{i=m+1}^ni}=\frac{n+1}{m}$

$\therefore\quad S=\frac{n+1}{m}$

For the given problem, $m=4$ and $n=7$ . Hence, $S=\dfrac{n+1}{m}=\dfrac{8}{4}=\boxed{2}$

I don't know how you got the 5 over 4 and the other ones, I was confused after that bit 😕

Wow. That's a complicated way of explaining adding fractions. It's simpler to just think that 1 is the same as a fraction that has the same number in the numerator and denominator. If that's confusing, just remember that any number (except 0) divided by itself equals one. So if you want to add 1 and 1/4, rewrite 1 as 4/4. Then add 4/4 plus 1/4 and you get 5/4.

Excellent !!