So I understand the method used in this solution, and I know my method is incorrect, but I was just looking for an explanation why.
I was thinking that if I choose any spot on the circumference, there's a $1/2^{n}$ probability that all $n$ points lie in the semicircle starting from that spot on the circumference. Why couldn't you just integrate this probability from 0 to 1 representing moving the spot around the whole circle?
Again, I know this is the wrong and just looking for reasoning why.
On
Your reasoning would be correct if you wanted the probability that all $n$ points are in a random semicircle. The desired probability, however, is the probability that all $n$ points all fall within the same (but not predetermined) semicircle.
Thus, for example, a single point obviously falls within the same semicircle as itself. However, integration gives the probability that the point falls within a randomly selected semicircle, which is $1/2$.
I don't see an obvious way to integrate something to get the right answer. Integrating is basically a form of addition (or (weighted) averaging if you prefer), but the things you're adding are not disjoint events.
E.g. let $E_\theta$ be the event that all the $n$ points are in the semicircle starting at the angle $\theta$ and going clockwise. Then for all $\theta, P(E_\theta) = 1/2^n$. But how do you add $E_\theta$ and $E_{\theta + \epsilon}$ for some small $\epsilon$? Those are overlapping semicircles. I find it hard to imagine any integral like $\int_0^{2\pi} P(E_\theta) \,\,\text{blah} \,\, d \theta$ would give you the right answer.
In particular, your original suggestion is equivalent to saying $\text{blah} = {1\over 2\pi}$, which is the density for picking the semicircle uniformly, so the result is $\int_0^{2\pi} P(E_\theta) {1\over 2\pi} d \theta = 1/2^n = $ Prob that a random semicircle (the one at a random $\theta$) contains all $n$ points (as Brian also pointed out).