How to prove that $\mathbb{C}_n$ is a subgroup of $(\Bbb C\setminus\{0\},\cdot)$. $\forall n \in \mathbb{N} \setminus \{0\}$  prove that $\mathbb{C}_n=\{z \in \mathbb{C}\;|\;z^n=1\}$ is a subgroup of $(\mathbb{C}\backslash\{0\},\cdotp)$ and that is a cyclic group of order $n$.
I know that a subgroup is when it's closed under the operation. Then to try to use the definition I noticed that $\mathbb{C}_1 = \mathbb{C}_{1+4n}$, $\mathbb{C}_2 = \mathbb{C}_{2+4n}$, $\mathbb{C}_3= \mathbb{C}_{3+4n}$, $\mathbb{C}_4 = \mathbb{C}_{4+4n}$, but I'm not sure if this is correct or not because I tried to prove it using induction but I failed. With that proven I thought that I could study the four sets one by one and see if they are subgroups to therefore prove that $\mathbb{C}_n$ is subgroup.

Subgroup of A means that it is a group and subset of A at the same time using the same operation.
So you need to show that (i) $\mathbb{C_n}\subset \mathbb{C}-\{0\}$ and (ii) $\mathbb{C}_n$ is a group using the same operation as in $\mathbb{C}-\{0\}$.
