Prove $\Pi_{k=1}^\infty(1+q^k)=\Pi_{k=1}^\infty\frac{1}{(1-q^{(2k-1)})}$ are equivalent generating series. 
Prove $\Pi_{k=1}^\infty(1+q^k)=\Pi_{k=1}^\infty\frac{1}{(1-q^{(2k-1)})}$ are equivalent generating series using a combinatorial proof.

The LHS of the equation is the generating series of partitions where distinct parts appear only once.  
The RHS looks similar to the equation for the generating series for all partitions which is $\Pi_{k=1}^\infty\frac{1}{1-q^k}$. In the original generating series, the exponent for $q$ would take the following values $\{1,2,3,4,\dots \}$. However, when we replace the exponent in the equation we are trying to prove, it takes the values $\{1,3,5,7,\dots \}$. The part I can't figure out is how this small change makes these equivalent with an explanation.  
 A: Hint: This is not combinatorial, but notice that
$$
1+q^k=\frac{1-q^{2k}}{1-q^k}
$$
Take the product and cancel.
A: It is known that the number of partitions of an integer into distinct parts (LHS) equals the number of partitions into odd parts (RHS).
For the various ways to prove it, see e.g. this excellent lecture, where at pag. 9 and 10 these two approaches are provided, which I am going to summarize herewith.


*

*conversion of the o.g.f.
$$
\begin{gathered}
  \text{o}\text{.g}\text{.f}\text{.}\;\text{DISTINCT} = \left( {1 + x} \right)\left( {1 + x^{\,2} } \right)\left( {1 + x^{\,3} } \right) \cdots  =  \hfill \\
   = \frac{{\left( {1 - x^{\,2} } \right)}}
{{\left( {1 - x} \right)}}\frac{{\left( {1 - x^{\,4} } \right)}}
{{\left( {1 - x^{\,2} } \right)}}\frac{{\left( {1 - x^{\,6} } \right)}}
{{\left( {1 - x^{\,3} } \right)}} \cdots  =  \hfill \\
   = \frac{1}
{{\left( {1 - x} \right)}}\frac{1}
{{\left( {1 - x^{\,3} } \right)}}\frac{1}
{{\left( {1 - x^{\,5} } \right)}} \cdots  = \text{o}\text{.g}\text{.f}\text{.}\;\text{ODD} \hfill \\ 
\end{gathered} 
$$

*Bi-jective correspondence
Given one partition of $n$ into distinct parts (i.e., sum of strictly decreasing sequence)
$$
n = d_{\,1}  + d_{\,2}  +  \cdots  + d_{\,m} 
$$
Consider that each integer can univocally be written as a power of $2$ (incl. $2^0$) multiplied by an odd integer
(factoring of less significant $0$s in the binary representation), so
$$
n = 2^{\,\alpha _{\,1} } O_{\,1}  + 2^{\,\alpha _{\,2} } O_{\,2}  +  \cdots  + 2^{\,\alpha _{\,m} } O_{\,m} 
$$
where the $O_j$ are odd numbers, possibly repeated (but in that case with a different $\alpha_j$).
We can then group for the different values of the odd numbers, as
$$
\begin{gathered}
  n = \left( {0 + 2^{\,\alpha _{\,1,1} }  + 2^{\,\alpha _{\,1,2} }  +  \cdots } \right)1 + \left( {0 + 2^{\,\alpha _{\,2,1} }  + 2^{\,\alpha _{\,2,2} }  +  \cdots } \right)3 + \left( {0 + 2^{\,\alpha _{\,q,1} }  + 2^{\,\alpha _{\,q,2} }  +  \cdots } \right)5 +  \cdots \quad \left| {\;\alpha _{\,q,j}  \ne \alpha _{\,q,k} } \right. =  \hfill \\
   = \mu _{\,1} 1 + \mu _{\,3} 3 + \mu _{\,5} 5 +  \cdots  \hfill \\ 
\end{gathered} 
$$
Therefore, each partition into distinct parts can be made to correspond bijectively to the binary representation of the $\mu$ vector,
hence to a partition into odd parts (repetition allowed).

A: (everything converges absolutely on the unit disk)
 $$\prod_{m=0}^\infty (1+z^{2^m})= \sum_{n=0}^\infty z^n = \frac{1}{1-z}$$
$$\prod_{k= 1}^\infty \frac{1}{1-q^{2k-1}} = \prod_{m\ge0,k\ge 1} (1+q^{2^m (2k-1)}) = \prod_{k = 1}^\infty (1+q^k)$$
