Showing $\prod_{g\in G}g=\prod_{g\in G}(a\cdot g)$ To put this in context, this is my first week abstract algebra. 
Let $(G,\cdot,e)$ a finite abelian group. And $a\in G$.
Show that:
$$\prod_{g\in G}g=\prod_{g\in G}(a\cdot g)$$
Here are my thoughts:
Let $G=\{g_1,...,g_n\}$. Then we obtain:
$$\prod_{g\in G}(a\cdot g)=ag_1\cdot...\cdot ag_n=a^n g_1\cdot...\cdot g_n=a^n\prod_{g\in G}g$$
I don't know how to show that $a^n$ must be $e$.
 A: Multiplication by $a\in G$ is an action on $G$, i.e. it is just a permutation of the elements in $G$. Since the group is abelian, permutation of the elements do not change the product hence the product remains the same.
Edit: As Javaman pointed out, using Lagrange's theorem seems simpler to follow from your work!
A: As JavaMan has 'hinted' in the comments, in any finite group $G$ with $|G| = n$, it is the case that $g^n = e$ for every $g\in G$.  But since you're only in your first week of abstract algebra, I'm not sure whether you have the tools (e.g. Lagrange's theorem) to prove that.
However, what you do know are the group axioms.  From these, you can prove that for any $a\in G$, the function $\phi_a: G\rightarrow G$ given by $\phi_a(g) = a\cdot g$ for all $g\in G$ is a bijection, and this allows you to prove the result.
A: Suppose that $G$ has $n$ elements. Consider the subset of $G$ of elements of the form $a^{-1}g$. If $a^{-1}g = a^{-1}g'$ then $g = g'$, so this subset has as many elements as all of $G$. Since $G$ is finite, this subset must be all of $G$. Thus you can write
$$\prod_{g \in G} a\cdot g = \prod_{a^{-1}g \in G} (a \cdot a^{-1} g)$$
Do you know how to finish the proof from here?
