Consider the ordinary generating function $e(x)=\sum_{n\geq0}e(n)x^n$ for the number $e(n)$ of partitions of the integer $n$ into even parts. Express $e(x)$ as a product of simple generating functions.
So I know that the number of partitions of $n$ into even parts is given by $$\prod_{i\geq0}\frac{1}{1-x^{2i}}$$ but I'm unsure how to express this as a product of simple generating functions, any help would be appreciated.
I agree with Peter Taylor's comment that the generating function you gave is pretty simple already. If you have to do something, you could use difference of squares to write $$ \prod_{i \ge 1} \frac{1}{1-x^{2i}} = \prod_{i \ge 1} \left(\frac{1}{1-x^i}\right)\left(\frac{1}{1+x^i}\right).$$
The more interesting thing is that the number of partitions of $n$ into even parts is a partition number. More precisely: Write $p(n)$ for the number of partitions of $n$. A partition with just even parts necessarily sums to an even number. Then $e(2n) = p(n)$ just by taking half of each part.
(Edited to reflect @DarijGrinberg's comment that $i$ should start from 1 rather than 0.)