Prove that the number of self-conjugate partitions of $n$ equals the number of partitions of $n$ into distinct odd parts First, I would love if someone can provide some clarification of this problem. Then possibly help me map out/begin a proof. 
So If I were taking the number $6$ and partitioning for example (just to make sure I understand what the question is asking):
The only partition with distinct odd parts would be $6=5+1$. However, for self-conjugate partitions I understand when I flip over the middle diagonal the picture should look exactly the same? That would also only happen once. 
How would I go about formulating a proof?

The Wikipedia article is quite good on a proof.
You can see that $x^{2n+1}=x^n\cdot x\cdot x^n$ to form both 'legs' of a self-conjugate partition in a Ferrers diagram.
Or, if you travel along the main diagonal and read only to the right, we are looking at the number of partitions into distinct parts, $\prod 1+x^k$. We need two of these - $\prod 1+x^{2k}$ - to form the reflection when travelling downwards, and we also need to supply the diagonal - $\prod 1+x^{2k}\cdot x$.
