Is this assumption "legal" when showing that the circle group is a subgroup of $\mathbb{C}^{\times} $? As an abstract algebra exercise, I need to show that the $n$th roots of unity is indeed a subgroup of $\mathbb{C}^\times$.
The only part of the proof I am concerned about is the multiplicative inverse. Here is what I have stated so far.

Let $a$ be an $n$th root of unity. Then $a^n = aa^{n - 1} = 1$. We
  will show that $a^{n - 1}$ is an $n$th root of unity. Observe that
  $(a^{n - 1})^n = a^{n^2 - n} = a^{n^2}a^{-n}$.

I want to say that, since $a^{-n}$ is a complex number, $a^{-n} = \frac{1}{a^n}$. Then $a^{n^2}a^{-n} = (a^n)^n\frac{1}{a^n} = 1^n\frac{1}{1} = 1$. Hence $a^{n - 1}$ is an $n$th root of unity and every $n$th root of unity has a multiplicative inverse.
It seems to me that I am making assumptions about how exponents work in an abstract setting and I am wondering if that is OK to do so?
 A: You're using all sorts of facts about exponentiation in $\mathbb C$ that are true, but you're right that they need to be proven beforehand. Specifically, you're using these:

For any $n, m \in \mathbb Z$ and $x\in\mathbb C$:
  $$x^{n+m}=x^nx^m$$
  $$(x^n)^m=x^{nm}$$

Note that $x^{-n}=1/x^n$ (where $n$ is positive) isn't true "because $x$ is a complex number", it's true because that's the definition of raising to a negative power. The fact that it's also true when $n$ is negative, however, is a theorem (that you aren't using here since your $n$ is positive).
You should prove the above facts yourself if you want to put your mind at ease. You'll need the fact that multiplication in $\mathbb C$ is associative and commutative, and you may find it easiest to separate all of them into different cases depending on the signs of $n$ and $m$. Oh, and $x^0=1$ is also just a definition, so you can assume that.
A: $a^{-1}$ is a $n$-th root of unit because $$(a^{-1})^n = (a^{-1})^n \cdot 1 = (a^{-1})^n \cdot a^n = (a^{-1}a)^n = 1^n = 1 $$ 
