The subfactorial function is defined as: $$!n = n!\sum_{i=0}^n\dfrac {(-1)^i} {i!}$$ I was curious and wanted to find out what $!0$ came out to be. Since I couldn't use it in the sum above, I used a different method by figuring out $!1$ first and using $!n = !(n-1)n-1$.
It turned out, that if you plugged in $0$ for this, you got $2$.
This seemed odd, so I used a reverse equation, $!n = \dfrac {!(n+1) +1} {(n+1)}$ and plugged in $2$ for $!0$ and it turned out correct.
This still seemed very odd, so I checked it on Wolfram Alpha, which say that $!0 = 1$
So what did I do wrong? And how exactly does $!0$ turn out to be $1$?
Using the definition you provided:$$ !1 = 1! \sum_{i=0}^1\frac{(-1)^i}{i!} = 1 \left(\frac{1}{1} + \frac{-1}{1}\right) = 0 $$Then putting this in the other recursive equation you gave:$$ !1 = !(1-1)(1) - 1 \implies 0 = !0 - 1 \implies !0 = 1 $$
Considering the definition provided on Wikipedia:
It's clear that the empty set has no elements to put into their previous position, likewise a set with only one element will always have to put that element in the first position.