Answer
$16y^{20}$
Work Step by Step
Using $(x^ay^b)^c=x^{ac}y^{bc}$ or the products-to-powers rule, the given expression, $
(-2y^5)^4
,$ simplifies to
\begin{array}{l}\require{cancel}
(-2)^{1(4)}y^{5(4)}
\\\\=
(-2)^{4}y^{20}
\\\\=
16y^{20}
.\end{array}