Given a group $G$ and given any element $g_i \in G$ does $g_i^n \in G$? Given a finite group $G$ and assuming $G = \{g_0,g_1,g_2,\dots,g_m\}$ for a positive $m$ value. 
If we choose a $g_i$ where $g_i \in G$.
Does the sequence $\{g_i,g_i^2,g_i^3 \dots \ \} \subset G$ due to the fact it is closed under binary operation? How do I guarantee that no element of the sequence will be bigger than $g_m?$ It must have some kind of 'circularity' to make things hold.
Can someone enlight me on how this statement holds(if holds) and why?
 A: This is part of the definition of a binary operation and a group.  The product of any two elements in a group is again an element of that group.
As for being "bigger than $g_m$", that doesn't make any sense.  There is no inherent notion of size in a group, so there is no bigger to speak of.  The indexes you've chosen to use to label your group elements are completely arbitrary and have nothing to do with the size of those elements.
It may help to work out a few examples of small finite groups.  Write down all their elements and work out how to compute products of those elements.
A: 
"due to the fact it is closed under binary operation? "

yes, is binary operation so $g_i^2\in g$ and $g*g_i^2 = g^3\in G$ and so by induction for all powers.

"How do I guarantee that no element of the sequence will be bigger than gm?"

Because that would be logically impossible.  As $G$ has only $m$ elements and all $g_i^k\in G$ there can only be at most $m$ (or fewer) of these $g_i^k$.

"It must have some kind of 'circularity' to make things hold."

Yes it must. And it does.
There must be infinitely many $g_i^k = g_i^l =g_i^w=.....$.
You will discover later that if $G$ is finite that there will be some value of $k$ where $g_i^k =e$, the identity element, and that therefore $g_i^{nk + r} = g_i^r$ and thus the infinite repeats.  The smallest value of $k$ so that $g_i^k =e$ is called the order of $g_i$.
[I haven't proven this is so.  But by the fact that there are infinitely many $n \in \mathbb N$ but only finitely many elements in $G$ there must be some $i,j\in \mathbb N, i < j$ but $g^i = g^j$.  From there we can prove that $e=(g^{-1})^i g^{i} = (g^{-1})^ig^{j} = g^{j-i}$.  So there is a $k:= j-i$ where $g^k = e$.]
......
But note.  If $G$ is infinite then all the $g_i^k$ can be distinct.
And BTW, please to not get confused be the subindexes.  They have nothing to do with anything.
If I had my way I'd have said the group, $G$,  has $m$ elements and not bother giving them labels until needed.
and the statement about powers of elements, I'd say as this: for any $g\in G$ we have $g^k \in G$.  (Which does mean that the set of $\{g,g^2,g^3,.....\}$ must be finite and the there must be (infinitely many) $i,k \in \mathbb N; i\ne k$ but where $g^i = g^k$.  (In other words "It must have some kind of 'circularity' )
I hope things are clear.
......
points to the OP for the phrase ""It must have some kind of 'circularity' " which is absolutely true.
And thanks to LSpice for telling me not to label the elements of $G$ as $\{e,a,b,c,....,z\}$ which just trades one confusion for another.
