Generating functions for partitions of n with an even number of parts and odd number of parts, and their difference. I've been trying to figure this out for more than 10 hours.
So far I have, for even number of partitions, $$P_e(x)=\sum_{k\ge1}(x^{2k}\prod_{i=1}^{2k}\frac{1}{1-x^i})$$ and for odd numbers
$$P_o(x)=\sum_{k\ge1}(x^{2k-1}\prod_{i=1}^{2k-1}\frac{1}{1-x^i})$$ 
Hopefully you can see the direction I'm headed in. According the the problem, the difference of the two
$$P_e(x) - P_o(x)$$ is supposed to equal $$\prod_{n\ge1}\frac{1}{1+x^n}$$
I have tried subtracting my results and cannot get anywhere. Can anyone shed some light on how to approach this problem?
 A: You don’t need your expressions for $P_e(x)$ and $P_o(x)$; just expand
$$\prod_{k\ge 1}\frac1{1+x^k}=\prod_{k\ge 1}(1-x^k+x^{2k}-+\ldots\;,$$
and notice that the individual $x^n=x^{k_1}x^{k_2}\ldots x^{k_m}$ terms of the outer product are positive or negative according as $m$ is even or odd, so the coefficient of $x^k$ in
$$\prod_{k\ge 1}\frac1{1+x^k}$$
is the number of partitions of $k$ with an even number of parts minus the number with an odd number of parts.
You may find this question and its answers of interest; it goes beyond your present problem, but I shouldn’t be surprised if you found yourself dealing with the topic soon.
A: Let $X$ be a set and $\mathcal P(X)$ be the set of all subsets of $X$. Furthermore, let $$P=\mathcal P(\Bbb Z_{\ge0}\times\Bbb Z_{\ge0}),$$
which is the set of all subsets $\{(n_1,k_1),(n_2,k_2),...,(n_j,k_j)\}\subset\Bbb Z_{\ge0}\times\Bbb Z_{\ge0}$. Finally, for any $\pi=\{(n_i,k_i):i\in I\}\in P$, define
$$||\pi||=\sum_{i\in I}n_ik_i.$$
Then for any uniformly convergent power series $f(q)=\sum_{n\ge0}a_nq^n$, we have
$$\begin{align}
\prod_{k\ge1}f(q^k)&=\prod_{k\ge1}\left(\sum_{n\ge0}a_nq^{nk}\right)\\
&=\sum_{\pi\in P}\prod_{(n,k)\in\pi}a_nq^{nk}\\
&=\sum_{\pi\in P}q^{||\pi||}\prod_{(n,k)\in\pi}a_n\\
&=\sum_{N\ge0}q^{N}\sum_{\,\,\,\pi\in P\\ ||\pi||=N}\prod_{(n,k)\in\pi}a_n.\tag1
\end{align}$$
The takeaway fact here is that the sum $\displaystyle \sum_{\,\,\,\pi\in P\\ ||\pi||=N}$ is being taken over all partitions of $N$.
In our case,
$$\prod_{k\ge1}\frac{1}{1+q^k}=\sum_{N\ge0}q^N\sum_{\,\,\,\pi\in P\\ ||\pi||=N}\prod_{(n,k)\in\pi}(-1)^n.$$
We can see that each partition $||\pi||=N$ can be re-written as
$$||\pi||=n_1k_1+n_2k_2+...+n_jk_j=\underbrace{k_1+k_1+...+k_1}_{n_1\text{ times}}+...+\underbrace{k_j+k_j+...+k_j}_{n_j\text{ times}}.$$
So the total number of parts in the partition $||\pi||=N$ is given by the quantity
$$\text{#}(\pi)=\sum_{(n,k)\in\pi}n=\sum_{i\in I}n_i.$$
Then we have
$$\prod_{(n,k)\in\pi}(-1)^n=(-1)^{\text{#}(\pi)}.$$
So, given a partition $||\pi||=N$, if there is an even number of parts (that is, $\text{#}(\pi)$ is even) then $(-1)^{\text{#}(\pi)}=1$. Likewise, if $\text{#}(\pi)$ is odd $(-1)^{\text{#}(\pi)}=-1$. Summing over all partitions $||\pi||=N$, we get
$$\begin{align}
\sum_{||\pi||=N}(-1)^{\text{#}(\pi)}=&\text{# of partitions with an even number of parts}\\
&-\text{# of partitions with an odd number of parts}\\
=& p_e(N)-p_o(N).
\end{align}$$
Therefore
$$\prod_{k\ge1}\frac{1}{1+q^k}=\sum_{n\ge0}(p_e(n)-p_o(n))q^n.$$
