Does (ab)^n = a^n b^n for 3 consecutive integers imply the group is abelian?

[natasha d]

Homework Statement

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

Homework Equations

for an abelian group G, ab=ba $\forall$a,b$\in$G
if a$\in$G, a has an inverse element also $\in$G 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

 

[Oster]

Your proof is a little hard to read. In your proof you seem to have proved ab=e for all a,b in G? that looks fishy...
this is a solution. https://www.physicsforums.com/showthread.php?t=529381

 

[natasha d]

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