The question what the smallest number in an empty set is might seem like nonsense because, you know, the set does not contain any numbers but this is just the kind of thing mathematicians worry about.
Before we start I want to take a look back at the last post where we used n! = n * (n-1)! to show 0! = 1. As is often the case in maths, there was actually a different way to see the result.
If n! Is the product of all numbers that are at least 1 but at most n then 0! has to be the product of no numbers at all. This again seems not to make a lot of sense but we can still figure it out.
When we have a product like x = a * b * c it is obvious that a * b * c * d = x * d for any number d. If we now take x to be the “empty” product we would still want this to hold so we get d = x*d for any d, implying x=1. Since 0! is such an empty product it also has to be 1.
We can use a similar technique for our minimum. If x is the minimum over some set, say {a, b, c}, and we add another element d, the new minimum of {a, b, c, d} will be d if x is greater than d and x otherwise.
If we add d to the empty set the new set will only contain d so d is also the smallest number in it. With x the minimum over the empty set this implies that x is greater than d. If this is supposed to be true for every number d then x has to be infinite. Note that while infinity isn't technically a number it still is a valid result here.
This might seem weird at first but this proof is really just a more exact way of saying that the minimum can only get smaller as we add elements so it should start out as big as possible.
I think this is a nice example of something often happening in maths where two seemingly unconnected problems turned out to be more similar than expected.
As a final note on factorials I want to add that the argument given above might also be used to justify something like (-1)! = 0 but this would violate the main condition 1 = 0! = 0 * (-1)!. In particular you might notice that this not possible for any value of (-1)!.