Does an inverse exist for a set of complex numbers with operation of multiplication? I will give some context of my question: I was trying to solve this problem

Let $S$ be a finite non-empty set of non-zero complex numbers which is closed under multiplication. Show that $S$ is a subset of the set $\{z \in \mathbb C: |z|=1 \}$. Show that $S$ is a group, and deduce that for some $n \in \mathbb N$, S is the set of n-th roots of unity.

When trying to prove that $S$ is a subset of $\{z \in \mathbb C: |z|=1 \}$, I tried to argue that since $S$ is closed, $\forall a \in S, \ \exists \ \text{a corresponding} \ n \in \mathbb N^+ $ such that $$a^n = a \tag{1}$$ and hence $|a|=1$.
Now I want to prove that the identity element is in $S$, can I argue from $(1)$ that $a^{n-1}=e=1$? 

Your reasoning is correct: for each $a \in S$, the finiteness of $S$ implies that there is an $n_a \in \Bbb N \setminus \{ 1 \}$ such that $a^ {n_a} = 1$. Since $S$ is closed under multiplication, it follows that $1 = a \cdot a \cdot \dots \cdot a \in S$. Finally, $a ^{-1} = a ^{n_a - 1} \in S$, so $S$ is a group.
