Let $A$ be a matrix having entries from a commutative ring. Then how can we prove that $A$ is invertible $\mod m$ for a positive integer $m$ if and only if $\det A$ is invertible $\mod m$?
Further if $A$ has entries from $\mathbb{Z}_m$, then can we also say that $A^n = 0$ for some positive integer $n$ implies $\det A^n \equiv 0 \mod{m}$?
EDITED TO ANSWER THE NEW QUESTION
Over any commutative ring $R$, the matrix $A\in$Mat$_n(R)$ is invertible if and only if $\det(A)$ is invertible in $R$:
a) If $A$ has inverse then $AA^{-1}=I_n$ implies $\det(A)\det(A^{-1})=1$, so $\det(A)$ is invertible.
b) If $\det(A)$ is invertible, as there exists the adjoint matrix adj$(A)$, which satisfies $$\text{adj}(A)A=\det(A)I_n,$$ we have that $\det(A)^{-1}$adj$(A)=A^{-1}$.
In addition, if $A^n=0$ then $\det(A^n)=(\det A)^n=0$, so if $A$ is nilpotent in Mat$_n(R)$ then $\det(A)$ is nilpotent in $R$, but not necessarily $0$ (unless there are no nontrivial nilpotents in $R$). The easiest example is a nilpotent element seen as a size-1 matrix (e.g., $2\in\mathbb{Z}_4$). More in general, if $m=p^k$, you can pick the diagonal $(k-1)\times(k-1)$ matrix $A$ with $p$ as diagonal entries, which has $A^k=0$ and $\det(A)=p^{k-1}\neq 0$, e.g. $$\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$$ in $\mathbb{Z}_8$.