0!=1???

0!=1

I could just say that it is define that way which is certainly true.

or I could say that 0! is the number of ways that 0 items can be odered - that would be just one way.

Now

5C1 is the number of ways that 1 item can be  chosen from 5 it equals      $$^5C_1=\frac{5!}{1!(5-1)!}=\frac{5!}{1*4!}=5\\\\$$

5C0 is the number of ways that 0 item can be  chosen from 5 it equals  [the answer has to be 1 there is only one way of chosing nothing]

    $$\\^5C_0=\frac{5!}{0!(5-0)!}=\frac{5!}{0!*5!}=\frac{1}{0!}\\\\
$ the only way this can work is if 0!=1$\\\\
so\\\\
0!=1$$

Here's an  informal "proof"....

3! =  4! / 4

2! =  3! / 3

1!  = 2!/ 2

So, by extension

0!   = 1! / 1    = 1