Find the last two digits of the number $9^{9^9}$ 
Find the last two digits of the number $9^{9^9}$ . 

[Hint: $9^9 \equiv 9 \pmod {10} $; hence, $9^{9^9}$ = $9^9+10k$ ;now use the fact that $9^9 \equiv 89 \pmod {100}$]

$$9^{9^9}=(10-1)^{9^9}\equiv(-1)^{9^9}+9^9\cdot10^1\cdot(-1)^{9^9-1}\equiv-1+10\cdot9^9\pmod{100}$$
Now, $\displaystyle9^9=(10-1)^9\equiv-1\pmod{10}\implies10\cdot9^9\equiv-10\pmod{10\cdot10}$
$$\implies9^{9^9}\equiv-1-10\pmod{100}\equiv100-11$$
