Why does 0! = 1? (Part 1)

This common question arises because 0! does not fit the usual definition for n!. Recall that, for positive integers, we have

5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120

4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24

3! = 3 \cdot 2 \cdot 1 = 6

2! = 2 \cdot 1 = 2

1! = 1

Going from the bottom line to the top, we see that start at 1, and then multiply by 2, then multiply by 3, then multiply by 4, then multiply by 5. To get 6!, we multiply the top line by 6:

6! = 6 \cdot 5! = 6 \cdot 120 = 720.

Because they’re formed by successive multiplications, the factorials get large very, very quickly. I still remember, years ago, writing lesson plans while listening to the game show Wheel of Fortune. After the contestant solved the final puzzle, Pat Sajak happily announced, “You’ve just won $40,320 in cash and prizes.” My instantaneous reaction: “Ah… that’s 8!.” Then I planted a firm facepalm for having factorials as my first reaction. (Perhaps not surprisingly, I was still single when this happened.)

Back to 0!. We can also work downward as well as upward through successive division. In other words,

5! divided by 5 is equal to 4!.

4! divided by 4 is equal to 3!.

3! divided by 3 is equal to 2!.

2! divided by 2 is equal to 1!.

Clearly, there’s one more possible step: dividing by 1. And so we define 0! to be equal to 1! divided by 1, or

0! = \displaystyle \frac{1!}{1} = 1.

Notice that there’s a natural way to take another step because division by 0 is not permissible. So we can define 0!, but we can’t define (-1)!, (-2)!, \dots.

In Part 2, I’ll present a second way of approaching this question.