Find The Last Two Digits Of $9^{8^7}$ Find the last two digits of $9^{8^7}$.
I tried finding a secure pattern for the last two digits of powers of $9$ but that didn't work.  Any answers?

Alternatively
By FLT 
$3^{\phi(100) = 40}= 9^{20} \equiv 1 \mod 100$
$9^{8^7} = 9^{2^{21}} \equiv 9^{2^{21} \mod 20} \mod 100$
$2^{\phi(5)=4} \equiv 1 \mod 5$ so $2^{21=4*5 + 1} \equiv 2 \mod 5$
$2^{21} \equiv 0 \mod 4$ os $2^{21} \equiv 2 + 5k \mod 20$ where $4|2 + 5k$. 
i.e. $2^{21} \equiv 12 \mod 20$.
so $9^{2^{21}} \equiv 9^{12}  \mod 100$
$9^{12} = (10 - 1)^{12} = 10^{12} - .... -12*10 + 1 \equiv -120 + 1 \equiv 81 \mod 100$
