Finite subgroup generated by an element

Question

$G$ is any group with identity element $e$ and $x\in G$. If $\langle x\rangle$ is of finite order $a$, why should $a=\min\{n\in\mathbb{N}^*\ |\ x^{n}=e\}$? In fact why should $\{n\in\mathbb{N}^*\ |\ x^{n}=e\}\ne\emptyset$? I am not able to derive a contradiction from the assumption $$(\forall n)(n\in\mathbb{N}^*\implies x^n\ne e)$$

Answer 1

Well if $\langle x\rangle$ has finite order then certainly the elements $x,x^2,\dots,x^k,\dots$ are not distinct, say $x^i=x^j$ for some $0<i<j$. Then $x^{j-i}=e$. Can you take it from there?