An infinite group $G$ and $\forall x\in G, x^n=e$ 
Let $G$ be an infinite group and $n\in \mathbb N$. If for any infinite subset $A$ of $G$ there is $a\in A$ such that $$a^n=e,~~~~(e=e_G)$$ then prove that for every element $x\in G$ we have $x^n=e$.

This question was asked me and honestly I could not find any way to break it well. I worked on the order of an element $x$ to show that every element is of finite order but it seemed not to complete a solid approach. Thanks for the time! 

Notice that $G$ has no elements of infinite order. Suppose there exists $y \in G$ with $|y| \nmid n$. Replacing $y$ with a power of $y$, we may assume $\text{gcd}(|y|, n) = 1$. If $y^G = \{gyg^{-1} \mid g \in G\}$ is infinite, we are done (since $|y| = |gyg^{-1}|$ for any $g \in G$). 
On the other hand, if $|y^G| = |G : C_G(y)| < \infty$, then $|C_G(y)| = \infty$, so there exists an infinite sequence of distinct elements $a_1, a_2, \ldots$ in $C_G(y)$ with $a_i^n = e$ for all $i$. But $|a_iy| = \text{lcm}(|y|, |a_i|) \nmid n$ for all $i$, so $\{a_iy \mid i = 1, 2, \ldots\}$ is an infinite set of elements, none of whose orders divide $n$.
