I found an equation accidentally when doing my research about branching processes. I think it is correct but I don't know how to prove it: \begin{equation} \sum_{n=1}^m \frac{1}{n!} \sum_{\sum_{i=1}^n m_i = m, \\ m_i \in \mathbb N_+} \frac{1}{m_1\cdots m_n} = 1, \quad m\in \mathbb N_+. \end{equation} I found it by enumeration by hand:
- $m=1$, $$\frac{1}{1!} \cdot \frac{1}{1} = 1.$$
- $m=2$, $$\frac{1}{1!} \cdot \frac{1}{2} + \frac{1}{2!} \cdot \frac{1}{1\cdot 1} = \frac{1}{2} + \frac{1}{2} = 1.$$
- $m=3$, $$\frac{1}{1!} \cdot \frac{1}{3} + \frac{1}{2!} \cdot \left( \frac{1}{1\cdot 2} \cdot 2 \right) + \frac{1}{3!} \cdot \frac{1}{1\cdot 1\cdot 1} = \frac{1}{3} + \frac{1}{2} + \frac{1}{6} = 1.$$
- $m=4$, $$\frac{1}{1!} \cdot \frac{1}{4} + \frac{1}{2!} \cdot \left( \frac{1}{1\cdot 3} \cdot 2 + \frac{1}{2\cdot 2} \right) + \frac{1}{3!} \cdot \left( \frac{1}{2\cdot 1\cdot 1} \cdot 3 \right) + \frac{1}{4!} \cdot \frac{1}{1\cdot 1\cdot 1\cdot 1} = \frac{1}{4} + \frac{1}{3} + \frac{1}{8} + \frac{1}{4} + \frac{1}{24} = 1.$$
- ...
I tried $m$ until 5 and I was surprised that the equation holds for all of them. I think this is not a coincidence. I also tried to prove it by induction, but I don't know how to apply the inductive hypothesis.
I feel that this equation should relate to integer partition and factorization, because the second summation is over partitions of $m$ and the product $m_1\cdots m_n$ is like a factorization of some integer.
But still, I have no idea to prove it. Have anyone seen it somewhere? Or does anyone know how to prove it? Any references or hints will be appreciated.