<s>[INST] Proving that odd partitions and distinct partitions are equal I am working through The Theory of Partitions by George Andrews (I have the first paperback edition, published in 1998).
Corollary 1.2 is a standard result that shows that the number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts.
Andrews's proof uses generating functions, but it contains this step that has me confounded:
$$
\prod_{n = 1}^\infty \frac{1 - q^{2 n }}{1 - q^n} = \prod_{n = 1}^\infty \frac{1}{1 - q^{2 n - 1}}.
$$
Can anybody point me in the right direction?
I tried turning $1 - q^n$ into an infinite sum and then expanding, to get
$$
\frac{1 - q^{2 n }}{1 - q^n} = 1 + q + q^2 + \dots + q^{2 n -1},
$$
but I don't see how this manipulation will help me.
 [/INST] See if this helps?
\begin{align*}
\prod_{n = 1}^\infty \frac{1 - q^{2 n }}{1 - q^n} &= \frac{1 - q^{2}}{1 - q}\cdot\frac{1 - q^{4 }}{1 - q^2}\cdot\frac{1 - q^{6 }}{1 - q^3}\cdot\dots\\
&= \left[\frac{1}{1-q}\cdot\frac{1}{1-q^3}\cdot\frac{1}{1-q^5}\dots\right]\left[\frac{1 - q^{2}}{1 - q^2}\cdot\frac{1 - q^{4 }}{1 - q^4}\cdot\frac{1 - q^{6 }}{1 - q^6}\cdot\dots\right]\\
&= \left[\frac{1}{1-q}\cdot\frac{1}{1-q^3}\cdot\frac{1}{1-q^5}\dots\right]\left[1\right]\\
&= \prod_{n = 1}^\infty \frac{1}{1 - q^{2 n - 1}}\\
\end{align*}
Where the second step indicates a rearrangement of the terms. Now one must be careful in infinite rearrangements -  you should check for the absolute convergence of an infinite sum. see here
You should check this for validity of the second step.
</s>