Is this a legitimate way to prove $0!=1$?

Question

if $n$ is any integer, we find that $\dfrac{n!/(n+1)}{(n-1)!/n}$ is $\dfrac{n^2}{n+1}$ through simplification. By putting in $n = 1$, we find that $\dfrac{1/2}{0!} = \dfrac{1}{2}$. Thus, we get $0! = 1$. Is this way of proving correct?

Answer 1

In your "proof", $n!$ is only defined when $n≧１$. So, $\frac{n!/（n+1)}{(n-1)!/n} $ is defined only when $n-1≧１i.e. n≧２$.

So you cannot put $n＝１$ into your identical equation.

If you want to define $0!$ naturally , natural requirement shall be $n!＝n\cdot（n-1)!$,take n＝１,then we have no choice but to define $0!＝1$. This is a standard definition of $n!（n≧０,n∈\mathbb{Z}）$ but not a proof of $0!＝１$.

Answer 2

No because $$\dfrac{n!/(n+1)}{(n-1)!/n} = \frac {n^2}{n+1}$$ only if $n\not = 1$

You are using it for $n=1$ as well.