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!
 A: 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 $100$.
A: Since $7$ and $100$ are coprime with respect to each other, you can use Euler's theorem:


*

*$7^{\phi(100)}\equiv1\pmod{100}$

*$\phi(100)=\phi(2^2\cdot5^2)=2\cdot4\cdot5=40$

*$7^{40}\equiv1\pmod{100}$

*$7^{2008}=7^{40\cdot50+8}=(7^{40})^{50}\cdot7^8$

*$(7^{40})^{50}\cdot7^8\equiv1^{50}\cdot7^8\equiv1\cdot5764801\equiv5764801\equiv01\pmod{100}$

A: 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?
A: Just a clarification to Subhadeep's answer in case it is confusing.
Let us first build a small table with everything larger than 100 in parentheses:
$$\begin{array}{lr}7^0 = &1\\
7^1 =& 7\\
7^2 =& 49\\
7^3 =& (3)43\\
7^4 =& (24)01\\
\end{array}$$
So we see that after 4 we get "back" to 1. This means $7^{4k}$ will end in 01 for all $k$. So if we can calculate the exponent modulo 4 we can see which of 01,07,49,43 it will be. 
As the exponent 2008 is divisible by 4 it must be "01".
A: $$7^4\equiv 01\pmod {100}\\\implies (7^4)^{502}\equiv (01)^{502}\pmod {100}\\\implies 7^{2008}\equiv 01\pmod {100} $$.
Hence the last two digits are $01$.
