Suppose dividing with [math]![/math] was possible (which is where your proof falls short). Given then that [math]n! = n(n-1)!\\[/math] we would get:[eqn]
\dfrac{n!}{!} = n = n(n-1)
[/eqn]
It is immediately evident that this is not true for all [math]n \in \mathbb{N}_{>0}[/math], neither is it true for [math]n = 1[/math] since [math]1 \ne 1(1-1)[/math].
[math]\text{Q.E.D.}[/math]
[math]\dfrac{n!}{n}=(n-1)! \implies 1!=\left(1-1\right)!=0![/math]
