# How do you simplify #mn^2p^0?

Jun 15, 2015

$m {n}^{2} {p}^{0} = m {n}^{2}$
For any base, $b$,
${b}^{0} = 1$
So $m {n}^{2} {p}^{0} = m {n}^{2} \cdot \left(1\right) = m {n}^{2}$