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

$9^{\phi(100)}=9^{40}\equiv 1\pmod{100}\implies9^{40k}\equiv 1\pmod{100}$
Now consider $9^{16}=40k+?$ , that is the same as finding $x\equiv 9^{16}\pmod{40}$
$9^{\phi(40)}=9^{16}\equiv 1\pmod{40}\implies 9^{16}=40k+1$
$9^{40k}\equiv 1\pmod{100}\implies 9^{40k+1}\equiv 9\pmod{100}$
