Are infinite products commutative? While reading a textbook, I came across the following proof (for integer partitions into odd parts and distinct parts):

The following steps can be justified by taking finite products and then passing to the limit: 
$\begin{align}
  \frac{1}{1-q}\frac{1}{1-q^3}\frac{1}{1-q^5} \ldots &= \frac{1}{1-q}\frac{1-q^2}{1-q^2}\frac{1}{1-q^3}\frac{1-q^4}{1-q^4}\frac{1}{1-q^5}\frac{1-q^6}{1-q^6}\ldots\\
  &= \frac{1-q^2}{1-q}\frac{1-q^4}{1-q^2}\frac{1-q^6}{1-q^3}\ldots\\
  &= (1+q)(1+q^2)(1+q^3)\ldots
  \end{align}$

I'm sure that the above reasoning is fine, but I can't help but have a problem with the equality from the first line to the second.
To me, this raises the following question(s):


*

*Are infinite products commutative?

*If not, how might the above proof be justified using limits?

 A: One may go at this problem in two ways:
(i) Consider the occurring expressions $1\pm q^r$ as analytic functions of the complex variable $q$ with $|q|\ll1$. Using the fact that
$$\bigl|{\rm Log} (1+z)\bigr|<{3\over2}|z|\qquad\bigl(|z|<{1\over2})$$
one can rewrite your identities as identities about absolutely convergent series. It follows that they are true as identities about certain analytic functions of $q$.
(ii) One may work in the realm ${\cal P}$ of formal power series (see, e.g., the first chapter of Henrici's Applied and computational complex analysis, I). Here ${1\over1-q}$ is an "abbreviation" for the series $1+q+q^2+\ldots$. Any such series is a bona fide element of ${\cal P}$, and an infinite sum or product of such series converges to a limit $a_0+a_1 q+a_2q^2+\ldots$, if each individual coefficient $a_r$ can be computed in only finitely many operations. To be exact: When
$$a_0^{(n)}+a_1^{(n)} q+a_2^{(n)}q^2+a_3^{(n)}q^2+\ldots\qquad(n\geq1)$$
is a sequence of such series in ${\cal P}$ then this sequence converges to the series
$$b_0+b_1q+b_2q^2+b_3q^3+\ldots$$
in ${\cal P}$ iff for each $r\geq0$ there is an $n_0=n_0(r)$ such that
$$a_r^{(n)}=b_r\qquad\forall n\geq n_0\ .$$
It is easily seen that this is the case in your formulae, as later summands or factors only involve higher powers of $q$.
