I want to show
\begin{align} \prod_{n\geq 1} (1+q^{2n}) = 1 + \sum_{n\geq 1} \frac{q^{n(n+1)}}{\prod_{i=1}^n (1-q^{2i})} \end{align}
I know one proof via self-conjugation of partition functions with Young Tableaux. But it seems not natural for me. [In the process of the proof it appears Durfee square, etc]
Is there any other (simple?) proof for this equality?
Substituting $\sqrt{q}$ for $q$, we get $$ \prod_{n=1}^{\infty}(1+q^n)=\sum_{k=0}^{\infty}{\frac{q^{\binom{k+1}{2}}}{\prod_{i=1}^{k}{(1-q^i)}}}. $$
The left-hand side is the generating function for partitions with distinct parts (a part of each size $n$ occurs $0$ or $1$ times).
On the other hand, if a partition with distinct parts has $k$ parts, then the $i$th smallest part is of size at least $i$. Given a partition $0<\lambda_1<\lambda_2<\dots<\lambda_k$, subtract $i$ from the size of the $i$th smallest part to get a partition with parts $0\le\lambda_1-1\le\lambda_2-2\le\dots\le\lambda_k-k$ with $\le k$ parts (after exluding the $0$s), whose conjugate is a partition with the largest part $\le k$. The "staircase" we subtracted has $1+2+\dots+k=\binom{k+1}{2}$ cells. That yields the summand on the right-hand side: $$ q^{\binom{k+1}{2}}\prod_{i=1}^{k}{\frac{1}{1-q^i}}. $$
I think this is about as simple as it gets. The Durfee square need not be involved, as you can see.
