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 $9^9 \equiv 89 \pmod {100}$]

Our observations start like this;
$9=-1 ($mod $10) \Rightarrow 9^9=(-1)^9 ($mod $10)$ or $9^9=-1 $(mod $10)$ 
We have  $-1=9 ($mod $10)$, thus $9^9=9 ($mod $10)$    ...i
This statement directly implies that upon dividing $9^9$ by $10$ we get a remainder $9$.Thus $9^9=9+10k$ 
Also,$9^9=89 ($mod $100) \Rightarrow 9*9^9=9*89 ($mod $100)$
or $9^{10}=801($mod $100)$. Also, $801=1 ($mod $100)$ thus making
$9^{10} = 1 \pmod{100}$       ....ii
$(9^{10})^k=1^k ($mod $100)$   from ii
$9^9*(9^{10})^k=9^9 ($mod $100)$ or, $9^{(9+10k)}=9^9 ($mod $100)$. Since,$9^9=89 ($mod $100)$, we have $9^{(9+10k)}=89 ($mod $100)$
This statement implies that upon division of $9^{9^9}$ by $100$ we get a remainder of $89$. Thus last two digits are $89$.
