# 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.