Find the sum of the last 2 digits of 9^103

What is the product of the last 2 digits of 2004^2004

What other ways to do these besides Euler's theorem and finding the pattern?

if you use factorials for (a-b)^103 you can write the equation back to front and add it together to get twice the value but cancelling out the intermediate factorials then divde the equation by half to get the number itself then just add the last two digits of it.

Someone explain it?