For any $n>1$ , $ Aut(\mathbb Z^n) \cong GL(n , \mathbb Z)$ ? And for any $m,n >1$ , $Aut(\mathbb Z_m ^n) \cong GL(n , \mathbb Z_m)$ ?

Question

Is it true that for any $n>1$ , $ Aut(\mathbb Z^n) \cong GL(n , \mathbb Z)$ ?

And is it true that for any $m,n >1$ , $Aut(\mathbb Z_m ^n) \cong GL(n , \mathbb Z_m)$ ?

( One problem I'm having with showing these automorphisms is the part that I cannot show that for any automorphism $f$ on $\mathbb Z^n$ , or $\mathbb Z_m^n$ , the matrix $\Big [ f(e_1) \space f(e_2) \space ... \space f(e_n)\Big]$ is invertible , where $e_i$ is the ordered pair , written in column form , whoose $i$-th position is $1$ and all other position is $0$ )