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!

Start by listing some cases of $7^n \mod 100:$
$$7^0=\color{red}{01}$$
$$7^1=\color{red}{07}$$
$$7^2=\color{red}{49}$$
$$7^3=3\color{red}{43}$$
$$7^4=24\color{red}{01}$$
$$7^5=168\color{red}{07}$$
$$7^6=1176\color{red}{49}$$
$$7^7=8235\color{red}{43}$$
$$7^8=57648\color{red}{01}$$
$$7^9=403536\color{red}{07}$$
$$7^{10}=2824752\color{red}{49}$$
$$7^{11}=19773267\color{red}{43}$$
$$7^{12}=138412872\color{red}{01}$$
$$\dots$$
Do you see the pattern?
