Prove its abelian is this proof correct

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Homework Statement


(ab)^{n}= a^{n}b^{n} for any 3 consecutive numbers n \inN

Homework Equations


for an abelian group G, ab=ba \foralla,b\inG
if a\inG, a has an inverse element also \inG such that aa^{-1} = e


The Attempt at a Solution


doesnt look right but here's the attempt

http://pics.livejournal.com/jackdnutter/pic/000013d1
 
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Oster said:
that looks fishy...

yeah it does, which is why I asked. Whats wrong with it though? if someone could point out the flawed reasoning maybe it'd better my understanding of group theory.
Thanks for the link, the (ba)^{∞} proof was hilarious
 
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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