If A is some invertible $n \times n$ matrix then show $\det(A^n) = (\det(A))^n$ for all $n\in \mathbb{Z}$

Question

So there exists $A^{-1}$. I am assuming $\det(AB)=\det(A)\cdot\det(B)$ and $(A^d)^f=(A^{df})$

I know the proof for $\det(A^{-1})=(\det(A))^{-1}$ is:

$\det(I_n)=1$

$\det(A\cdot A^{-1})=1$

$\det(A)\cdot \det(A^{-1})=1$

$\det(A^{-1})=\frac1{\det(A)}=(\det(A))^{-1}$

But I'm not sure how to prove it when it's $n\in\mathbb{Z}$, and that if A is not invertible then for all $n\in \mathbb{N}$

Answer 1

I am assuming $\det(AB)=\det(A)\cdot\det(B)$ and $(A^d)^f=(A^{df})$

By your assumption, $$ \det(A^1)^n = \det(A^{1 \cdot n}) $$