Partitions into distinct even summands and partitions into (not necessarily distinct) summands of the form $4k-2,k\in\Bbb N$ 
Prove that the number of ways to partition $n\in\Bbb N$ into distinct even summands is equal to the number of ways of partitioning $n$ into (not necessarily) distinct summands of the form $4k-2,k\in\Bbb N$.

My thoughts:
It appeared to me, if $$n=\sum_{k=1}^pa_k,$$ where $a_1,\ldots,a_p$ are distinct even numbers, then $$\frac{n}2=\displaystyle\sum_{k=1}^pb_k, \quad b_k=\frac{a_k}2,$$ and $b_1,\ldots,b_p$ should be distinct summands.
On the other hand, $$n=\sum_{k=1}^r c_k,$$ where $c_k=4m_k-2,m_k\in\Bbb N,$ so $$\frac{n}2=\sum_{k=1}^rd_k,d_k=\frac{c_k}2=2m_k-1,m_k\in\Bbb N,$$ so we might use the fact that the number of ways to partition $n\in\Bbb N$ into distinct summands is equal to the number of ways to partition $n$ into odd summands, but this doesn't seem convincing enough.
I thought we could compare the generating functions $$f_1(X)=\prod_{k=1}^\infty(1+x^{2k})=\prod_{k=1}^\infty(1+x^{4k-2})(1+x^{4k})$$ and maybe $f_2(X)=\displaystyle\prod_{k=1}^\infty\frac1{1-x^{4k-2}},$ but I couldn't get any further.
I also checked this answer about partitions of $n$ where no part is divisible by $d$ and partitions of $n$ in which no part occurs more than $d-1$ times, hoping it could help.
How can I approach this problem?
 A: 
We obtain
\begin{align*}
\color{blue}{\prod_{k=1}^{\infty}\left(1+x^{2k}\right)}
&=\prod_{k=1}^{\infty}\frac{\left(1+x^{2k}\right)\left(1-x^{2k}\right)}
{\left(1+x^{k}\right)\left(1-x^{k}\right)}\tag{1}\\
&=\prod_{k=1}^{\infty}\frac{\left(1+x^{2k}\right)\left(1-x^{2k}\right)}
{\left(1+x^{2k}\right)\left(1+x^{2k-1}\right)\left(1-x^{2k}\right)\left(1-x^{2k-1}\right)}\tag{2}\\
&=\prod_{k=1}^{\infty}\frac{1}
{\left(1+x^{2k-1}\right)\left(1-x^{2k-1}\right)}\tag{3}\\
&\,\,\color{blue}{=\prod_{k=1}^{\infty}\frac{1}
{1-x^{4k-2}}}\\
\end{align*}
and the claim follows.

Comment:

*

*In (1) we use $1-x^{2k}=(1+x^k)(1-x^k)$.


*In (2) we use $\prod_{k=1}^{\infty}\left(1+x^k\right)=\prod_{k=1}^{\infty}\left(1+x^{2k}\right)\left(1+x^{2k-1}\right)$.


*In (3) we do some cancellation.
A: I think I have the answer using generating functions:
As mentioned before, if $p_1(n)$ is the number of partitions of $n\in\Bbb N$ into distinct summands, and $p_2(n)$ number of partitions of $n\in\Bbb N$ into odd summands, then their respective generating functions $f_1(X)=\displaystyle\prod_{k=1}^\infty(1+X)$ and $f_2(X)=\displaystyle\prod_{k=1}^\infty\frac1{1-X^{2k-1}}$ are equal.
As in my post, $g_1(X)=\displaystyle\prod_{k=1}^\infty(1+X^{2k})$ is the generating functions for partitions of $n\in\Bbb N$ into distinct even summands and $g_2(X)=\displaystyle\prod_{k=1}^\infty\frac1{1-X^{4k-2}}$ is the generating function for partitions of $n\in\Bbb N$ into summands of the form $4k-2.$
Let $h(X)=X^2$
$$g_1(X)=\prod_{k=1}^\infty(1+(X^2)^k)=(f_1\circ h)(X)\\g_2(X)=\prod_{k=1}^\infty\frac1{1-(X^2)^{2k-1}}=(f_2\circ h)(X).$$
Now, $f_1=f_2\implies g_1=f_1\circ h=f_2\circ h=g_2.$
