Is it possible to prove $g^{|G|}=e$ in all finite groups without talking about cosets? Let $G$ be a finite group, and $g$ be a an element of $G$. How could we go about proving $g^{|G|}=e$ without using cosets? I would admit Lagrange's theorem if a proof without talking about cosets can be found.
I have a proof for abelian groups which basically consists in taking the usual proof of Euler's theorem and using it in a group, I do not know if it can be modified to work in arbitrary finite groups.
The proof is as follows: the function from $G$ to $G$ that consists of mapping $h$ to $gh$ is a bijection. Therefore 
$\prod\limits_{h\in G}h=\prod\limits_{h\in G}gh$ but because of commutativity $\prod\limits_{h\in G}gh=\prod\limits_{h\in G}g\prod\limits_{h\in G}h=g^{|G|}\prod\limits_{h\in G}h$.
So we have $\prod\limits_{h\in G}h=g^{|G|}\prod\limits_{h\in G}h$.
The cancellation property yields $e=g^{|G|}$.
I am looking for some support as to why it may be hard to prove this result without talking about cosets, or if possible an actual proof without cosets.  
Thank you very much in advance, regards.

Let me take a shot-
Let $o(g)=n$ for some arbitrary $g \in G$, then $g^n=e$ (and $n$ is least such positive integer), now if suppose $g^{|G|}\neq e$, then there exist there exist $t \in \mathbb{Z}$  which is also greater than $1$ such that $g^{|G|t} = e$, but then by division algorithm $\exists \ $ $q,r \in \mathbb{Z}$ such that $|G|t=nq+r$ and $0\leq r <n \implies g^{nq+r}=g^r=e$ $\implies$ $r=0$ $\implies$ $|G| = \frac{nq}{t} \implies g^{|G|}=g^{n(q/t)} \neq e $  (by hypothesis)   but $g^n=e$. 
Now the question is why does $t$ has to divide $q$, but I argue as (avoiding order of element divides order of $G$, which is the question itself) that it must divide, as once $g^n=e$, then raising $e$ to the power $\frac{q}{t}$ doesn't make sense if $t$ does not divide $q$. 
