Incorrect. Remember that $\displaystyle{{n}}^{{-{{a}}}}=\frac{{1}}{{{n}}^{{a}}}={{\left(\frac{{1}}{{n}}\right)}}^{{a}}\usepackage{cmbright}\Huge\$ . What is $\displaystyle\frac{{1}}{{{8}}^{{2}}}\usepackage{cmbright}\Huge\$ in simplified form?