How to determine $\prod_{g\in G}g$? It’s an exercise in my textbook.

Let $G$ be a finite Abelian group, then determine
  $$\prod\limits_{g\in G}g.$$

Actually, I do not quite get what it is asking. What does it mean by “determine”? What needs to be determined?
Moreover, there’s a following task asking me to show with the help of my first question that 
$$(p−1)! \equiv −1~{\rm mod}~p~~ (p~{\rm prime}).$$ I’m then getting more confused...
Any help, hint or detail, would be greatly appreciated. Thanks!
PS: They are exercises from my textbook, on page 49 of The Theory of Finite Groups, An Introduction (page 62 of the pdf).
 A: Every element $g \in G$ has an inverse $g^{-1}$. There are two cases:
$(i)$ $g=g^{-1}$, in which case $g$ is either the identity or an element of order $2$.
$(ii)$ $g \ne g^{-1}$.
In the product $\prod\limits_{g\in G}g$ each $g$ in case $(ii)$ can be paired with its inverse. This leaves the product of elements in case $(i)$.
When $G$ is $(\mathbb{Z}/p\mathbb{Z})^\times$ then
$\prod\limits_{g\in G}g = \prod\limits_{k=1}^{p-1}k \mod p=(p-1)! \mod p$
and each element in the product can be paired with its inverse apart from $1$ and $p-1$.
A: The answer of your first question has been pointed out by Kiryl Pesotski. 
For your second question, notice that $(p-1)!$ is the product $\prod_{g \in G} g$ where $G$ is $(\mathbb{Z}/p\mathbb{Z})^\times$, and that $-1$ is of order $2$ in this group.
In this type of exercise, determine means reduce the product to an element you know ($e$ here) or to an element which has specific properties, that you must find.
A: Not a complete answer
A lot of your question is already answered by the links Kiryl Pesotski posted. The second question is already answered here and also in the links posted. There remains one case, however. For your first question:


*

*If there is no element of order 2, every element and its inverse appear in the product, and the identity $e$ appears once, such that the product equals $e$. By Cauchy's theorem, this is the case when $\mathrm{order}(G)$ is odd.

*If $\mathrm{order}(G)$ is even, again by Cauchy's theorem, there is an element of order $2$. Suppose there are $k$ elements of order $2$ and denote these by $g_1,g_2,\ldots,g_k$. Then, because $G$ is abelian by assumption, $\{e,g_1,g_2,\ldots,g_k\}\subset G$ is a subgroup (verify this). Then by Cauchy's theorem again, the order of this subgroup must be even, thus $k$ is odd. If we write out $\prod_{g\in G}g$ now, we observe that for all $g\not\in\{e,g_1,g_2,\ldots,g_k\}$, both the element and its inverse appear exactly once in the product, thereby yielding the identity element. Thus the product reduces to $\prod_{i=1}^{k}g_i$. For $k=1$, it's simple, $\prod_{g\in G}g=g_1$, for $k=3$ also: $\prod_{g\in G}g=g_1\circ g_2\circ g_3=g_3^2=e$, since $g_1\circ g_2\not\in \{e,g_1,g_2\}$. For any $k>3$ (the general case) the product reduces to exactly one element in $\{e,g_1,g_2,\ldots,g_k\}$ but to be honest I don't see how to infer anything about this.
So this is not a duplicate of the linked questions, the way I see it, because you are asked to determine the product in all generality. Maybe someone else knows how to calculate the product for $k>3$.
