Dear Uncle Colin,

Why does $0! = 1$ and not 0?

- Nothing Is Logical

Hi, NIL, and thanks for your message!

My best explanation for this - by which I mean, the one I can get some people to accept, goes like this:

- $4! = 4 \times 3 \times 2 \times 1 = 24$.
- To get to $3!$, you divide $4!$ by 4 and get $3! = 6$.
- To get to $2!$, you divide $3!$ by 3 and get $2! = 2$.
- To get to $1!$, you divide $2!$ by 2 and get $1! = 1$.
- And to get to $0!$, you divide $1!$ by 1 and get $0! = 1$.

(You can’t go any further, because you’d have to divide by zero - but you can extend the factorial function into non-integers using the gamma function.)

A similar argument works for powers:

- $3^2 = 3 \times 3 = 9$.
- To get to $3^1$, you divide $3^2$ by 3 and get $3^1 = 3$.
- To get to $3^0$, you divide $3^1$ by 3 and get $3^0 = 1$.

(You *can* continue this pattern into the negative numbers!)

In both cases, when you multiply nothing together, you get 1. (We say “1 is the multiplicative identity”).

It feels strange that the value of “no things” is 1 if you’re multiplying all of the no things together, and 0 if you’re adding them up, but it has to be that way for maths to work.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Barney Maunder-Taylor

A similar argument applies to the sequence 3^3^3^3, 3^3^3, 3^3, 3, then what? And after that??? In each case, do log base 3 to get from one term to the next. The next term is therefore … ?

## Colin

Nice — although I suspect someone asking this question may not be entirely comfortable with logs 😉