The question changes to $(21{)}^{289}\times 4^{289}$

Using the last two digits technique for 1 -

For all numbers ending in 1, last digit is always 1.

Second last digit is the product of ${10}^{th}$ digit of the base and the unit digit of the power.

E.g.

In this question ${21}^{289}=\_\_81$

Using the last two digits technique based on the pattern of 2, we can arrive at the answer very fast.

$4^{289}=(2^2{)}^{289}=2^{578}=(2^{10}{)}^{57}\times 2^8=\_\_24\times \_\_56$

Since, $(2^{10}{)}^{odd}$ power always ends in 24.

Answer $=\_\_81\times \_\_24\times \_\_56=\_\_64$