Consider a nilpotent matrix A of index of nilpotency n , then ${A}^{n}=O$ and ${A}^{n-1}\ne O$ where O denotes null matrix of same order as of A

Now ${A}^{n}=O\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}det(A{)}^{m}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}det(A)=0$

but ${A}^{n-1}\ne O\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}det(A)\ne 0$

I know I am wrong in the second statement because a matrix not being equal to a null matrix doesn't imply that it has to be necessarily non-singular.

Therefore aren't we allowed to take determinant on both sides for the inequation $A\ne O$ ?

Now ${A}^{n}=O\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}det(A{)}^{m}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}det(A)=0$

but ${A}^{n-1}\ne O\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}det(A)\ne 0$

I know I am wrong in the second statement because a matrix not being equal to a null matrix doesn't imply that it has to be necessarily non-singular.

Therefore aren't we allowed to take determinant on both sides for the inequation $A\ne O$ ?