Eliminate Some Powers Immediately

We will find the last four digits, hoping for some bonus points ;)

We can write the given number as $9^{8^{7^{4n}}}$ for some large integer $n$ . Now $7^4=(50-1)^2\equiv 1 \pmod{100}$ so that $7^{4n}\equiv 1 \pmod{100}$ as well. Since $\phi(125)=100$ , we can conclude that $8^{7^{4n}}\equiv 8 \pmod{125}$ . This congruence holds modulo 8 as well, so that $8^{7^{4n}}\equiv 8 \pmod{1000}$ . The Carmichael's lambda function of 10000 is 500, and we can conclude that $9^{8^{7^{4n}}}\equiv 9^8 \equiv \boxed{6721}\pmod{10000}$ .

We will denote $9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}$ as $x$

$9^{400} \equiv 1 \pmod{1000}$

We have to find ${8^{7^{6^{5^{4^{3^{2^{1}}}}}}}} \pmod{400}$

We can use Chinese remainder theorem.

We can write the given number as $8^{7^{4n}}$ for some large integer $n$

$8^{7^{4n}} \pmod{25} \Rightarrow 8^{(7^{4n}) \pmod{20}} \pmod{25} \Rightarrow 8 \pmod{25}$

( $7^4 \pmod{20} \Rightarrow 1 \pmod{20}$ )

$8^{7^{4n}} \equiv 0 \pmod{16}$

Applying Chinese remainder theorem and simplifying,

We get:

$x \pmod{1000} \Rightarrow 9^8 \pmod{1000} \Rightarrow 1-10\binom{8}{1}+100\binom{8}{2} \pmod{1000} \Rightarrow 721 \pmod{1000}$