Proof verification: For all natural numbers, either $\sqrt{n}$ is irrational or $n$ is a perfect square

Question

I am attempting problem 19 from Chapter 2 of Spivak's Calculus book. It says the following:

A fundamental theorem about integers, which we will not prove here, states that [prime factorization] is unique, except for the order of the factors. Thus, for example, $28$ can never be written as a product of primes one of which is $3$, nor can it be written in a way that involves $2$ only once (now you should appreciate why 1 is not allowed as a prime).

Using this fact, prove that $\sqrt{n}$ is irrational unless $n=m^2$ for some natural number $m$.

My attempt goes as follows:

Let $n\in\mathbf{N}$ and suppose that $\sqrt{n}$ is rational. Then there are $a,b\in\mathbf{Z}$ with $b\neq 0$ such that $\sqrt{n}=a/b$. Note that $n>0$, so $a,b$ must be natural number. Then:

$n=a^2/b^2$

$n=\large\frac{(a_1\cdot a_2 \cdot ... \cdot a_{k_1})^2}{(b_1\cdot b_2 \cdot ... \cdot b_{k_2})^2}$

where $a_1,...,a_{k_1}$ and $b_1,..., b_{k_2}$ are the unique prime factors of $a$ and $b$ respectively. Since $n$ is a natural number then $k_2\leq k_1$ so $a/b$ must also be a natural number. Therefore, we can conclude that $n$ is a perfect square.