What is the connection between Euler's Formula and solving differences of powers? Euler's formula is the following: $e^{ix} = \cos(x) + i\sin(x)$
By difference of powers, I mean $a^n - b^n = 0$ specifically $x^n - 1 = 0$
When I calculate $x^3 - 1 = 0$ I get $x = (1, \frac{-1}{2} +\frac{i \sqrt3}{2},\frac{-1}{2} -\frac{i \sqrt3}{2})$ This is similar to cos and sin relating to these numbers. Specifically $-\frac{1}{2} + \frac{i\sqrt{3}}{2} = \cos(\frac23\pi) + i\sin(\frac23\pi) = e^{\frac23\pi i}$. My question is "Why do these relate?"
 A: Think about the geometry of the complex plane.
To multiply two complex numbers, you multiply the lengths and add the angles. That tells you that the solutions to
$$
x^n -1
$$
form a regular $n$-gon on the unit circle at angles $2k\pi/n$. The exponentiation rule tells you those points are the points $e^{2k\pi i/n}$.
You can see the real and imaginary parts of those points in Euler's formula.
As @soupless suggests, look up roots of unity.
A: On the request of @aschepler, I'm turning my comment into an answer with a bit of extra detail.
From what I understand, either the question is asking: why is $e^{i\theta} = \cos \theta + i\sin \theta$? Or it's asking: Given that $e^{i\theta} = \cos \theta + i\sin \theta$, why are the solutions of $x^n - 1 = 0$ of the form $e^{i\left(\frac{2\pi k}{n}\right)}$. I'll answer both.
Starting with the first question. Consider the function
$$f(\theta) = \frac{\cos\theta + i\sin\theta}{e^{i\theta}} = \left(\cos \theta + i \sin \theta\right)e^{-i\theta}$$
We know immediately that $f(0) = 1$.
We can also tell that
$$f'(\theta) = (-\sin \theta + i\cos \theta)e^{-i\theta} + (-i\cos \theta + \sin \theta)e^{-i\theta} = 0$$
Since $f'(\theta) = 0$ everywhere, $f(\theta)$ must be constant. Since $f(0) = 1$, $f(\theta) = 1$. So
$$\left(\cos \theta + i \sin \theta\right)e^{-i\theta} = 1 \\
\implies \cos \theta + i \sin \theta = e^{i\theta}$$
Now the next question. Notice that $1 = \cos(2\pi k) + i\sin(2\pi k) = e^{i(2\pi k)}$ for every integer $k$. In the equation $x^n = 1$, we can subtitute this result to get:
$$x^n = e^{i(2\pi k)}$$
Taking the $n$th root on both sides gives us the desired result:
$$x = e^{i\left(\frac{2 \pi k}{n}\right)}$$
