Given $(A-I)^2 = 0$, can we say det$(A)=1$ and tr$(A)=n$?

Let $\lambda$ be an eigenvalue (in the complex numbers) of $A$; then $Av=\lambda v$ for some $v\ne0$ in $\mathbb{C}^n$; therefore $$ (A-I)v=(\lambda-1)v $$ and $$ 0=(A-I)^2v=(A-I)(\lambda-1)v=(\lambda-1)^2v $$ Therefore $(\lambda-1)^2=0$ and so $\lambda=1$. Thus $A$ has just the eigenvalue $1$, with algebraic multiplicity $n$.

The determinant of $A$ is the product of the eigenvalues and the trace their sum (counted with their multiplicity).