Find the last two digits of the given number Problem:

Find the last $2$ digits of $7^{2008}$.

Unfortunately I have no idea how to solve this problem. I know that for the last digit, we have to apply the concept of cyclicity, but I'm not aware of how to extend this to the last $2$ digits. I would be truly grateful for any help. Many thanks in advance!

Hint: $~7^{2008}=49^{1004}=(50-1)^{1004}.~$ Now expand using the binomial theorem, and notice that all terms except for the first two are multiples of $50^2$, and therefore of $100$.
