Answer
See below
Work Step by Step
Assume $A=diag(a_1,...a_n)\\
B=diag(b_1,...b_n)$
then $(AB)_{ij}=\sum a_{ik}b_{kj}\\
=a_{ii}b_{ij}\\
\rightarrow AB=diag(a_1b_1,...a_nb_n)$
Let $A=B=D$ and $D=diag(\lambda_1^k,\lambda_2^k,...\lambda_n^k)$
we have $D^k=diag(\lambda_1^k,...,\lambda_n^k)$
