Let $G = \{g_1, g_2, \dots, g_n\}$ be a finite abelian group, prove that for any $x ∈ G$, the product $$xg_1 \cdot xg_2 \cdot \cdot \cdot xg_n = g_1 \cdot g_2 \cdot \cdot \cdot g_n.$$
I can easily see why $g_1 \cdot g_2 \cdot \cdot \cdot g_n = e$, this follows from the group being abelian and having an inverse for every element. But I can't see why $x^n = e$. We haven't looked at many propositions yet. What I have so far is that if $\operatorname{ord}(x) = k$, that I need to prove $n \bmod k = 0$, from which will nicely follow that $\operatorname{ord}(x)$ divides the group order.
Edit: changed some notation mistakes in the product.