Physical substance/meaning of $0!=1$? I know the question sounds silly, however all I could find is the mathematical proof justifying the same but the convincing inference is still missing.
My intent basically was to ask for the better comprehension of "there is only one way of arranging no objects." 
Please be polite.
 A: Number of permutations of 0 elements...?
A: There are a couple of explanations.


*

*It's useful because it extends the equality $(n+1)! = n!(n+1)$ up to $n=0$

*The $\Gamma$ function, which is equal to $\Gamma(n) = (n-1)!$ for all $n\geq 2$, has a value of $1$ at $1$, so if we define $n! = \Gamma(n-1)$, then $0! = \Gamma(1) = 1$

*$n!$ is the number of ways you can order a set with $n$ elements, and there is exactly one way to order an empty set.

*As one of the commenters said, $0!$ is also the empty product and thus equal to $1$.

A: Intuitively, the factorial of an integer $n$ is the "product of all the integers from $1$ until $n$" - so $1!=1$, $3!=1\cdot2\cdot3=6$, etc., as you know. If we use this definition for $0$, then we get what is called an "empty product" - the product of no numbers.
What should the empty product be? First, let's look at addition, since it's easier. When you take the sum of no numbers, the empty sum, it should be the additive identity: zero. This is clear when you consider sums such as
$$\sum_{\substack{(2k+1) \text{ is even} \\ k \,\in\, \mathbb{N}}} k$$
Since there are no such $k$, this sum is the empty sum, which is zero.
Similarly, the empty product should be the multiplicative identity, which is $1$. Thus $$0!=1$$
A: I think that $0!=1$ by definition: $(n+1)!=(n+1)\cdot n!$ needs a start value and this value is the only integer that makes $1!=1$, which make sense in combinatorics.
