Question about a proof of FTA in A classical Introduction to modern number theory I just started to work on A classical introduction to modern number theory by K. Ireland and M. Rosen and I have a question.
In the first chapter, they prove the following lemma:

Every nonzero integer can be written as a product of primes.

Then they mention the following.

By collecting terms we can write $n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$, where the $p_i$'s are primes and the $a_i$'s are nonnegative integers. [...] The exponents $a(p)$ are nonnegative integers and of course $a(p)=0$ for all but finitely many primes.

Then after proving some other lemmas, he proves that $a(p)=\text{ord}_pn$ (where $\text{ord}_pn$ is the largest nonnegative integer $t$ for which $p^t\mid n$). 
My question is: isn't he (either Rosen or Ireland) assuming already (implicitly of course) that $ a(p)=\text{ord}_pn$ when he says that we can collect terms and write $ n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$? (or he's just proving that $\text{ord}_pn$ is unique?)
I understand that given a prime $p,$ $\text{ord}_p$ is a function that maps integers to nonnegative integers (and $+\infty$ if the input is $0$), and that this implies that the prime factorization of a nonzero integer $n$ is unique (up to order), but only if we "collect terms" as the authors do, but how do we exactly define the $a(p)$'s?
When we prove (the "usual" proof that everyone knows) the uniqueness part of the FTA by induction on $n$ (in this book they also prove the existence by induction), we don't need to "collect terms", so, we do not really assume anything "illegal".
I hope that my question is clear and I really appreciate any help (and I'm sorry if this is a silly question, maybe I'm just a bit confused).

A similar question may come up again, so I will turn my comment into an answer. 
Given any representation of $n$ as a product of primes, we can certainly collect equal primes and produce a representation in "exponential" form. That says nothing about the essential uniqueness of the representation, it is just a book-keeping move, since we are taking commutativity of multiplication  for granted.
