<s>[INST] 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?
 [/INST] Since \begin{align}(10-1)^{10}&=\sum_{i=0}^{10} \begin{pmatrix} 10 \\i \end{pmatrix}10^i(-1)^{10-i}\\&=(-1)^{10}+10(10)(-1)^9+\sum_{i=2}^{10} \begin{pmatrix} 10 \\i \end{pmatrix}10^i(-1)^{10-i} \\&\equiv1\mod 100\end{align}
We have 
$$9^{10} \equiv 1 \mod 100$$
Since $8^7 \equiv (-2)^7 \equiv -128 \equiv 2 \mod 10,$
$9^{8^7}\equiv9^2 \equiv 81 \mod 100$
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