Last two digit of number raised to exponents. 
Find the last two digits of $3^{3^{2014}}$.

Attempt: 
 First I try to work with $3^{2014}$. So we can work on $\text{mod 10}$.
Then, $$\begin{align}&3^1 \equiv 3\pmod{10}\\ &3^2 \equiv 9 \pmod{10}\\ &3^3 \equiv 27 \equiv 7\pmod{10}\\ &3^4 \equiv 81 \equiv 1\pmod{10}\end{align}$$
So  $3^{2014} = 3^{(4\cdot503) + 2} \equiv 3^0 \cdot 3^2 \equiv 9 \pmod{10}$.
So the last digit is 9. However I am ask to find the last two digit. Can anyone please help me? I would really appreciate all the help.

Hint: by Euler's totient theorem, $3^{40} \equiv 1 \pmod{100}$. Alternatively, note (by playing with a calculator) that $3^{20} \equiv 1 \pmod{100}$.
