$\alpha \nmid n \Rightarrow \alpha$ is a generator of group $G$ Let $G=\left \langle x \right \rangle$ a cyclic group s.t. $ord(x)=n$. Let $a=x^\alpha$ s.t. $\alpha \nmid n$ and $0<\alpha<n$ . Prove that $x^\alpha$ is a generator of $G$.
My partial answer: Let $A:=\left \langle x^\alpha \right \rangle$. We'll assume in counterwise the oposite. Then, $|G|=n$ thus $|A|\leq n$. $|A|\ne n$ because elsewise we would get $\left \langle x^\alpha \right \rangle = G$. Thus, $|A|<n$. I would like for some help from here or to get another direction.

That is false. 
Counter-example:
The multiplicative group $\;\bigl(\mathbf Z/11\mathbf Z\bigr)^{\times}$ is cyclic of order $10$, generated by the congruence class of $2$. However, $x=2^4$ has order $5$ (a divisor of $10$): modulo $11$,
$$a=5,\enspace a^2=3,\enspace a^3=4,\enspace a^4=9,\enspace a^5=1.$$
The correct assertion is thatit is true if and only if $\gcd\alpha,n)=1$. More generally, if $x$ generates a cyclic group of order $n$, $a=x^\alpha$ generates a cyclic group of order 
$$\frac n{\gcd(\alpha,n)}$$
