Help clarifying a question regarding (i think) cyclic groups

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Homework Statement


Let G be a group with a finite number of elements. Show that for any a in G, there exists an n in Z+ such that an=e.

Homework Equations


a hint is given: consider e, a, a2,...am, where m is the number of elements in G, and use the cancellation laws.

The Attempt at a Solution


so part of the trouble I'm having (i'm guessing the most important part), is figuring out what they're asking. is the number n I'm looking for a number that when any a in G is raised to, it gives the identity? i.e. (i'm just picking random letters) an= e and bn=e when a and b are not equal but n is the same in both? i was working on this question for a while thinking it to mean that a particular ap might have n0 while aq might have n1 to take it to e (identity).
 
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and then another problem in the same set as the previous one, that may hinge on the same language:
show that if G is a finite group with identity e and with an even number of elements, then there is a not equal to e such that a*a=e.
does this mean every element in G is its own inverse, or just one particular element? not really sure where i'd begin either way, but just to point me in the right direction.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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