Does the definition of factorial fail for $0!$

Question

How does one understand $0!=1$. Is it purely a theoretical value for convenience? as the definition of factorial i.e., $n!=n\times(n-1)\times(n-2)\times...1 $ actually fails here. How can it be logically understood by the definition of factorial.

Answer 1

It is somewhat for convenience. One reason is that we can think of the factorial as satisfying the recurrence relation $(n+1)! = (n+1)n!$, which if we extend the definition to $n=0$ then it is only compatible if we have $$1 = 1! = 1\cdot 0!$$ which means that we have to take $0! = 1$.

Answer 2

If you want to look for a logical way why $0! = 1$. Don't look at the definition, but at what it means in other fields. A factorial is also the amount of ways you can arrange a certain amount of objects. So $0!$ can be seen as the amount of ways you can arrange $0$ objects, which is in only $1$ way.

Answer 3

Count backwards: $$(n-1)! = \frac{n!}{n}$$ $$0! = \frac{1!}{1} = \frac{1}{1} = 1$$

Answer 4

Another approach is $n! = \prod_{i=1}^{n} i$. Now similarly to: the sum of all elements of the empty set is the additive neutral element ($\sum_{i\in\emptyset} i = 0$) we can say that the product of all elements of the empty set is the multiplicative neutral element ($\prod_{i\in\emptyset} i = 1$).

Taking this to account we receive $$0! = \prod_{i=1}^{0} i = \prod_{i\in\emptyset} i = 1$$