Product of all elements in finite abelian group times random group element

I'm just starting a course on group theory and the question in struggling with is as follows:

Let $$G$$ be a finite Abelian group and let $$x\in G$$. Proof: $$\prod_{g\in G}xg=\prod_{g\in G}g$$

I've tried a couple things, since the group is Abelian we have: $$\prod_{g\in G}xg=x^{\left|G\right|}\prod_{g\in G}g$$. I know for a fact that $$x^{\left|G\right|}=e$$ however this is only proved using Lagrange's Theorem which is still about 5 weeks away and I can't use right now. So my question is either:

1. Is there another way, that is without Lagrange's Theorem, to prove that $$x^{\left|G\right|}=e$$?
2. Do you have any tips for solving this question?

If the elementns of $$G$$ are $$g_1,g_2,g_3,\ldots, g_n$$, then $$xg_1, xg_2, \ldots, xg_n$$ are the same elements, just possibly in a permuted order.
• I am sorry but you should justify your answer by the fact that $g \to xg$ is bijective because it possesses an inverse which is $g \to x^{-1}g$. – Jean Marie Feb 21 at 8:20