The fundamental theorem of arithmetic simply states that every positive integer has a unique prime factorization, which you use in your proof. So it's really nothing special, and your proof seems normal to me.
Here's a slightly shorter way of putting it that might be a bit too succinct:
Suppose for contradiction that there is a natural n not a perfect square with sqrt(n) rational. Then there exist natural a,b such that b2n = a2. A natural number is a perfect square if and only if the exponents in the prime factorization of the number are all even. Since b2 is a perfect square and n is not, there is an odd exponent in the prime factorization of n and therefore there is an odd exponent in the prime factorization of b2n, since an odd exponent is added to an even one. But then b2n is not a perfect square, while a2 is, a contradiction.
EDIT: I am reminded of an overkill proof that 21/n is irrational for all natural n > 1. For the case n = 2 use the proof above. For n > 2, suppose that 21/n = b/a is rational. Then 2 = bn/an, and so 2an = bn. Well, in this case, an + an = bn would be a contradiction of fermat's last theorem, so 21/n cannot be rational.