A number theory problem by Omar Nader

We can denote $n!$ as $\frac{(n+1)!}{n+1}$

for example: $3!$ = $\frac{4!}{4}$ = $\frac{4 \cdot 3 \cdot 2 \cdot 1}{4}$ = $3 \cdot 2 \cdot 1 = 6$

Because of this concept, we can denote $0!$ as $\frac{1!}{1}$ or $\frac{1}{1}$ which can be simplified as 1

Mathematicians have found that $0!$ is much more useful being defined as 1, rather than 0. So, the factorial $0!$ is defined as 1. This can be seen when we use such formulas as $_nC_k = \dfrac{n!}{k!(n -k)!}$ in instances where $n = k.$

For instance, when given a list of 5 toppings for a pizza, how many 5 topping subsets are there? Well, there's just one! But let's see how that works in the formula $_nC_k.$ Here, we have 5 toppings to choose from, so $n = 5.$ Furthermore, we wish to see how many 5 topping subsets there are, so $k = 5.$ Therefore, our answer is

$_5C_5 = \dfrac{5!}{5!(5 - 5)!} = \dfrac{5!}{5!0!} = \dfrac{\cancel{5!}}{\cancel{5!}0!} = \frac{1}{0!}$

We know that there can only be 1 possible selection, so $0!$ must be 1 in order for $\frac{1}{0!}$ to return a value of 1.

Because it is

4! = 24, 3! = 6, 2! = 2, 1! = 1. If you see these numbers you can see a specific pattern. 3! = 24/4 = 6, 2! = 6/3 = 2, 1! = 2/2 = 1. Following this pattern, 0! = 1/1 = 1.