Let $p_0^d (n)$ be the number of partitions of n into distinct odd parts. Show that $$p(n) \equiv p_0^d(n) \pmod 2$$
So I have to show that the total number of partitions has the same parity as the number of partitions of n into distinct odd parts. I know that there are equal amounts of even and odd partitions, but other than that don't really know how to approach this.