Why does the definition of the Euler's number not violate the rule against division by zero?

Question

e= appears to be defined as the sum of the series 1/n! as n goes from zero to infinity.

But this implies that the first term is 1/0! which appears to violate the rule against division by zero

Answer 1

$0! = 1$

It is defined to be so.

You can also think about it in terms of what factorial means combinatorially.

$2!$ is the number of ways of placing 2 objects which is 2 ways.

$1!$ is the number of ways of placing 1 object, there is only 1 way.

And so what does $0!$ mean? Well there is only 1 way to place no objects!

Answer 2

This is one of those comment confusions that sometimes takes a while to understand. We define empty products to be equal to $1$, because we want the behavior to be induction - that is, we want: $$\prod_{i=1}^{n+1} a_i =\left(\prod_{i=1}^n a_i\right)\cdot a_{n+1}.$$

When $n=0$, then, $\prod_{i=1}^0 a_i$ is defined to be $1$. The coincides with defining $x^0=1$, for example (and it is why a lot of mathematicians define $0^0=1$.)

Similarly, we define empty sums to be zero. In the case of sums, that feels more natural and intuitive.

Answer 3

$0!$ is $1$. Why?

Well, $3!$ is the number of ways to arrange three objects: $$(1,2,3),(1,3,2),(2,1,3),\\(2,3,1),(3,1,2),(3,2,1)$$ So $3!=6$.

$2!$ is the number of ways to arrange two objects: $$(1,2),(2,1)$$ So $2!=2$.

$1!$ is the number of ways to arrange one object: $$(1)$$ So $1!=1$.

$0!$ is the number of ways to arrange zero objects: $$()$$ So $0!=1$.

Answer 4

$0!$ is defined as $1$, so there is no division by zero.