Last 2 digits of $\displaystyle 2014^{2001}$ How to find the last 2 digits of $2014^{2001}$? What about the last 2 digits of $9^{(9^{16})}$?

Hint:
For finding last two digits you need to reduce this modulo $100$. That is you ne need to find
$$2014^{2001} \equiv ? \pmod{100}.$$
This is the  same as asking
$$14^{2001} \equiv ? \pmod{100}.$$
Now in order to facilitate computation, you need to use Euler's Theorem. But keep in mind that $\gcd(14,100) = 2.$ So you need to adjust things a bit.
