Why $x^{n}=1$'s solution is equal to regular polygon in complex plane?

Very nice question. Here is the thing about complex multiplication: it works best in polar coordinates.

Every complex number $z$ can be written in the form $r e^{i \phi}$ where $r$ is a real number and $\phi$ an angle (so technically also a real number). When drawing $z$ in the complex plane the number $r$, called the modulus, is the distance from 0 to $z$ and $\phi$, called the argument is the angle that the line from 0 to $z$ makes with the positive real axis. In other words: $(r, \phi)$ are the location of $z$ in standard polar coordinates.

Now here is the thing you need to know:

If you multiply two complex numbers you multiply their moduli and add their arguments

So if $z = re^{i \phi}$ and $w = s e^{i \theta}$ then $zw = rse^{i (\theta + \phi)}$.

This has a very interesting consequence for numbers with modulus 1.

Multiplying an arbitrary complex number $w$ with $e^{i \phi}$ looks graphically just like rotating $w$ over angle $\phi$ around the origin (counter clockwise)

This simple result (much much much more simple than any part of Galois theory) is what you need to get your answer.

To find the $n$ points of a regular $n$-gon you start with the number 1. You rotate it $2\pi/n$ degrees to get to the next point on you n-gon, which corresponds to multiplying by $e^{i 2 \pi/n}$. Then you multiply it with that number again to get to the third point etc.

After $n$ steps you are back where you started: at the number 1. So apparently multiplying 1 $n$ times with $e^{i 2 \pi/n}$ yields $1$ again or in other words $(e^{i 2 \pi/n})^n = 1$ or, equivalently $e^{i 2 \pi/n}$ is a solution to $x^n = 1$.

Now in a completely similar fashion you can show that the other points on the $n$-gon are also solutions to that equation.

EDIT: what I showed above is that the points of a regular $n$-gon that

a) lie on a circle with radius 1 and

b) have number 1 as one of the points

all are solutions to $x^n = 1$. You asked the oposite question: why do the solutions of $x^n$ form a regular $n$-gon. What we know from the above is that at least some $n$-element subset of the set of solutions to $x^n = 1$ do form an $n$-gon, namely the $n$ points on the special $n$-gon discussed above.

So the only thing left two show is that the equation $x^n = 1$ has no other solutions besides the $n$ we just studied.

[END OF EDIT, original answer continues below]

There are two ways of doing that: you use the description of multiplication above to see that every other point in the complex plane has modulus either too small or too big or otherwise the wrong argument, or you use some abstract theory to show that no degree $n$ polynomial can have more than $n$ roots.