for any $k \in \mathbb{Z}^{+}$, how do we know $\omega^1$ is a primitive root of $x^k - 1$? I know that it is definitely a root.  The set $\{1, \omega, ..., \omega^{k-1}\}$ are the roots of unity.  But how are we guaranteed that it even exists and that furthermore it is a primitive root?
There is no need for any arguments specific to $ \mathbb{C} $ to see this. Let $ K $ be any field, and consider the polynomial $ X^n - 1 $. We first prove that if $ \alpha $ is a root, then its order must divide $ n $. Indeed, if $ k $ is the order of $ \alpha $, then we can divide $n $ by $ k $ to get $ n = kq + r $. Then, we have $ 1 = \alpha^n = \alpha^{kq + r} = \alpha^r $ where $ 0 \leq r &lt; k $, so by definition of order we have $ r = 0 $.

We will need another result: let $ \alpha $ and $ \beta $ be primitive $ m $ and $ n $th roots of unity, respectively, where $ \gcd(m, n) = 1 $. Then $ \alpha \beta $ is a primitive $ mn $th root of unity. To see this, note that if $ \alpha^k \beta^k = 1 $, we may take $ m $th powers to obtain $ \beta^{mk} = 1 $, which means the order $ n $ divides $ mk $, and therefore $ k $. Repeating the same argument shows that $ k $ is divisible by $ mn $, showing the result. Note that this extends easily to the product of arbitrarily many numbers whose orders are pairwise coprime.

Now, let $ n = p^k $ be a prime power. Every root of $ X^n - 1 $ then has some order of the form $ p^i $ where $ 0 \leq i \leq k $, and there are less than or equal to $ p^i $ elements of order $ p^i $ by Lagrange's theorem. However, we have that

$$ \sum_{i=0}^{k-1} p^i = \frac{p^k - 1}{p-1} \leq p^k - 1 &lt; p^k $$

so there must be some element $ \alpha $ with order $ p^k $, which is a primitive $n $th root of unity by definition.

To finish, let $ n = p_1^{q_1} p_2^{q_2} \ldots p_k^{q_k} $, and note that we may find $ \alpha_1, \alpha_2, \ldots, \alpha_k $ which are primitive $p_i^{q_i}$th roots of unity. By our second result,

$$ \alpha = \prod_{i=1}^{k} \alpha_i $$

then has order $ n $, finishing the proof.
It is because 1 is always coprime to k. If you looked at $\omega^i$ then it will be a primitive root iff $i$ is coprime to $k$. Think about generators of cyclic groups.