According to this $0!$ Should be not defined, but why it is not? According to my knowledge factorial means that the number is multiplied by the numbers preceding it up to $1$ . That's it .
$5!=5×4×3×2×1=120$
$4!=4×3×2×1=24$
$3!=3×2×1=6$
$n!=n×(n-1)×\cdots \times 2×1$
And similarly
$0!=0×(-1)×(-2)\times\cdots\times1$
But this is not possible as we have crossed 1 before and another 1 is not possible . So it should be ND.
Please tell me where I am wrong in this?

My favorite is the following. We define $n!$ to be the number of ways $n$ distinct objects can be ordered. Now, how many ways can you order $0$ objects? Well, you can either say that (1)"you can't", (2) "there are zero ways", or (3) "there is one way to organize zero objects". If you choose (1) you're just being defiant. Moreover, you would like to avoid zeroes as much as possible, since factorials do indeed appear in denominators so you try to find a way around (2). So we go with definition (3), because we choose to come to the agreement that there is one way to order zero objects.
