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

By Euler's Theorem
$2014^{\phi(25)}=2014^{20}\equiv 1\pmod{25}\implies2014^{2000}\equiv 1\pmod{25}\implies2014^{2001}\equiv 2014\pmod{25}\implies2014^{2001}\equiv 14\pmod{25}$
Obviously $2014^{2001}\equiv 0\pmod{4}$
Let $2014^{2001}=25n+14\implies25n+14\equiv 0\pmod{4}\implies n\equiv 2\pmod{4}$
$2014^{2001}=25n+14=25(4k+2)+14=64+100k$
