partitions with odd parts is equal to partitions with distinct parts

Statement

The number of partitions of $n$ with distinct parts is equal to the number of partitions with odd parts.

n	$p_{d i s t} (n) = p_{o d d} (n)$
0	1
1	1
2	1
3	2
4	2
5	3

Proof

We look at the generating functions. For distinct parts, well if $λ_{i}$ are distinct, then the multiplicity has to be either 0 or 1 for each number. Thus,

\sum_{n \geq 0} p_{d i s t} (n) x^{n} = \sum_{m_{1}, m_{2}, \dots; m_{i} \in {0, 1}} x^{1 m_{1} + 2 m_{2} + \dots} = \prod_{i = 1}^{\infty} (x^{i \cdot 0} + x^{i \cdot 1}) = \prod_{i = 1}^{\infty} (1 + x^{i})

On the other hand,

\begin{aligned} \sum_{n \geq 0} p_{o d d} (n) x^{n} & = \sum_{(m_{1}, m_{3}, m_{5}, \dots)} x^{1 m_{1} + 3 m_{3} + \dots} \\ = \prod_{i odd} \frac{1}{1 - x^{i}} \\ = \prod_{i \geq 0} \frac{(1 - x^{2}) (1 - x^{4}) \dots}{1 - x^{i}} \\ = \frac{(1 - x) (1 + x) (1 - x^{2}) (1 + x^{2}) \dots}{(1 - x) (1 - x^{2}) (1 - x^{3}) \dots} \\ = \prod_{i \geq 0} (1 + x^{i}) \end{aligned}

as desired.