Consider the following identity:
[2.4] Powers of Two Identity (2.9)
$$
\begin{align}
p(n \mid \text{ parts in }\{1\}) = p(n \mid \text{ distinct parts in }\{1,2,4,8,...\})
\end{align}
$$
That is, the number of integer partitions made of "1"s is the same as the number of integer partitions made entirely from distinct parts from powers of two.
Example 1
Take \(n = 7\). The only partition we can make out of 1s is
$$
\begin{align}
7 &= 1 + 1 + 1 + 1 + 1 + 1 + 1
\end{align}
$$
Using powers of two, we also know that the only partition we can make is
$$
\begin{align}
7 &= 2^0 + 2^1 + 2^2 \\
&= 1 + 2 + 4
\end{align}
$$
A Bijection
The merging process we used in Euler’s identity will merge pairs of ones into twos and then into fours and so on until all the parts are distinct as follows
$$
\begin{align}
7 &= 1 + 1 + 1 + 1 + 1 + 1 + 1\\
7 &= 2 + 2 + 2 + 2 + 1 \\
7 &= 4 + 4 + 1 \\
9 &= 8 + 1
\end{align}
$$
The splitting process takes any power of two and splits it into two equal parts over and over again. Since the only power of two that is odd is \(1\), then the process continues until we get all 1s.
$$
\begin{align}
9 &= 8 + 1 \\
9 &= 4 + 4 + 1 \\
9 &= 2 + 2 + 2 + 2 + 1 \\
9 &= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
\end{align}
$$