Is there any flaw in the factorial domain defined? Because $n!=n(n-1)!$. and so on and directly $n=0$ I get $0!$ as $0$.

Question

I know I can write $n!=n(n-1)!$.

Now putting $n=1$, I can see $0! = 1$ but is there any other way or that's the way we defined it.

Answer 1

$n!=n(n-1)!$ holds only if $(n-1)!$ exists ... (but when $n=0$, it doesn't exist).

We know $1!=1$, right? Take your equation with $n=1$ to get $1! = 1\cdot 0!$. So if $0!$ exists, we conclude that $0!=1$.

Answer 2

The factorial is generally defined on $\mathbb{N}=\{0,1,2,3,\dots\}$. The recursive rule $n!=n\times (n-1)!$ works for $n>0$, but for $n=0$ it invokes $(-1)!$, which is outside the domain.

You could use this recursive rule as a definition, together with $0!=1$, but possibly more common is the definition:

$$ n!=\prod_{i=1}^ni=1\times 2\times 3\times \cdots\times n$$

With this definition, we need to think about $0!$ as the empty product. The empty product needs to be $1$, because multiplying by $1$ leaves a number unchanged. With the empty product as $1$, we can guarantee this (desirable) property of products will always hold:

$$ \prod_{x\in S\cup T}x=\prod_{x\in S}x\times\prod_{x\in T}x$$

(A similar argument shows the empty sum needs to be $0$.)