$\Aut((\ZZ/p)^n)$ as a matrix group

Write \[ V=(\ZZ/p)^n\cong \FF_p^n. \] Every group endomorphism of the elementary abelian \(p\)-group \(V\) is automatically \(\FF_p\)-linear, because for \(a\in\FF_p\) and \(v\in V\), \[ \phi(av)=\phi(v+\cdots+v)=a\phi(v). \] Hence \[ \Endo_{\Grp}(V)=\Endo_{\FF_p}(V)\cong M_n(\FF_p), \] and the automorphisms are exactly the invertible linear maps: \[ \Aut_{\Grp}(V)\cong \GL_n(\FF_p). \]

To count them, choose the columns of an invertible matrix successively. The first column can be any nonzero vector, giving \(p^n-1\) choices. Once \(j\) linearly independent columns have been chosen, their span has \(p^j\) elements, so the next column has \(p^n-p^j\) choices. Therefore \[ \left|\Aut_{\Grp}\bigl((\ZZ/p)^n\bigr)\right| =|\GL_n(\FF_p)| =\prod_{j=0}^{n-1}(p^n-p^j). \]