What are the last two digits of 7^2008?

What are the last two digits of 7^2008?

Solution:
7^1 = 07
7^2 = 49  
7^3 = 343   -> 49 times 7 is 343 => last two digits => 43
7^4 = 2401  -> 43 times 7 is 2401 => last  two digits => 01
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801 -> last two digits again 01

01 -> repeats after every 4 times

Now, 2008 mod 4 = 0

Hence, the last two digits of 7^2008 = 01