$\mathbf{SETUP}$
By rephrasing the question of counting derangements from
"how many permutations are there with no fixed points?"
to
"how many permutations have cycle types that are non-unity partitions?"
one can use the well-known formula (with many names, but often cited as Cauchy's formula) for the number of cycles of type $\lambda = (1^{\alpha_1}, 2^{\alpha_2}, ..., n^{\alpha_n})$
\begin{align} N_{\lambda} = \frac{n!} {1^{\alpha_1} 2^{\alpha_2} ... n^{\alpha_n} \times \alpha_1! \alpha_2! ... \alpha_n!} = \frac{n!}{\Lambda_\lambda} \end{align}
to write the number of derangements as the sum of all possible cycles of type defined by the non-unity partitions of $n$ (written $P_{n,1}$ here)
\begin{align} !n = \sum_{\lambda \in P_{n,1}} \frac{n!}{\Lambda_\lambda} = \sum_{\lambda \in P_{n,1}}\frac{n!}{2^{\alpha_2} ... n^{\alpha_n} \times \alpha_2! ... \alpha_n!} \end{align}
$\mathbf{RESULT}$
This can be equated exactly with the more typical form for the subfactorial
\begin{align} !n = n!\sum_{k=0}^{n} \frac{(-1)^k}{k!} \end{align}
The $n!$ factorials cancel, leaving the surprising equality that connects the alternating sum of fractional factorials directly to the set of non-unity partitions!
\begin{align} \sum_{k=0}^{n} \frac{(-1)^k}{k!} = \sum_{\lambda \in P_{n,1}} \frac{1} {2^{\alpha_2} ... n^{\alpha_n} \times \alpha_2! ... \alpha_n!} \end{align}
The reason for this equality is intuitive - they are counting the same things! - but the arithmetic form is very surprising.
$\mathbf{QUESTION}$
What 'explains' the surprising arithmetic nature of this equality?
I am not a mathematician by training so I do not know: is this relation well-known?
Does its connection to (non-unity) partitions have any further uses?
A research paper I found that directly talks toward this kind of link between derangement and non-unity partitions is:
https://www.hindawi.com/journals/jmath/2021/6023081/
But it doesn't make quite the same jump connecting it to Cauchy's formula, nor does it answer why the surprising arithmetic equality appears to work (even if the 'reason' is clear!).
(This question stems from this previous question of mine, and indeed the 'remarkable' result MacMahon quotes on page 114 of his Combinatory Analysis (1915) is indeed this surprising equality.)
$\mathbf{EXAMPLE \ (SURPRISING?)}$
Just to highlight the surprise, at least arithmetically, take the case $n=7$. The left hand side becomes
\begin{align} \sum_{k=0}^{7} \frac{(-1)^k}{k!} = \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} - \frac{1}{7!} = \frac{103}{280} \end{align}
The right-hand side, noting that $P_{7,1} = \{(2^1, 5^1), (2^2,3^1), (3^1,4^1), (7^1)\}$ is then
\begin{align} & \sum_{\lambda \in P_{7,1}} \frac{1} {2^{\alpha_2} ... 7^{\alpha_7} \times \alpha_2! ... \alpha_7!} \\ & = \frac{1}{2^1 5^1 \times 1!1!} + \frac{1}{2^2 3^1 \times 2!1!} + \frac{1}{3^1 4^1 \times 1!1!} + \frac{1}{7^1 \times 1!} \\ & = \frac{1}{10} + \frac{1}{24} + \frac{1}{12} + \frac{1}{7} \\ & = \frac{103}{280} \end{align}
Try this for all $n$, it works! It has to...
$\mathbf{EPILOGUE}$
A furtherance of this relation can be seen by the fact that the left hand side will always end in a $\frac{(-1)^n}{n!}$, and the right hand side will always contain a trivial partition $\lambda = (n^1)$, so always includes the term $+\frac{1}{n!}$.
Thus, for even $n$, these terms cancel; for odd $n$, they sum to form $+\frac{2}{n!}$ on the right hand side.
