Quote by TylerH
Is it enough, once given that, if ab^{n}, then a^{n}b[supn[/sup], to say that:
n^{1/2}=p/q
n=p^{2}/q^{2}
nq^{2}=p^{2} Contradiction if n is not expressible as k^{2}, where k is an arbitrary integer.
The contradiction being that there is an uneven power of a prime factor on one side, which is impossible for a squared number.

I don't know what you mean on the top there (what is a?), but yes, this is what all the other proofs are driving at in more or less lengthy ways... It is merely a matter of comparing prime factorizations.
This method of proof generalizes neatly to the n'th root case. Comparing the exponents of the primes in the respective factorizations will yield the wanted contradiction.