The magical number "8"

We know, $888=8×3×37$ .

Now, $\boxed{ 8^{88^{888}}\pmod{8}=0}$

Finding $8^{88^{888}}\pmod{3}$ ,

$8^{88^{888}}\equiv 2^{88^{888}} \equiv (-1)^{88^{888}} \equiv 1 \pmod 3 \\ \therefore \boxed{8^{88^{888}}\pmod 3=1}$

Finding $8^{88^{888}}\pmod{37}$ ,

$8^{\phi(37)}=8^{36}\equiv 1\pmod{36} \\ 88^{888} \pmod{36}$

Now, $36=4×9$ .

$88^{888}\equiv 0\pmod{4} \\ 88^{888}\equiv 7^{888} \pmod 9 \\ (-2)^{888}\equiv 8^{296} \equiv 1\pmod 9$

By applying Chinese Remainder Theorem to $x \equiv 0 \pmod 4 \\ x\equiv 1 \pmod 9$ We will get,

$88^{888} \pmod {4×9=36}=28 \\ 8^{88^{888}}\equiv 8^{28} = (8^3)^9\cdot 8 \pmod{37} \\ 31^9\cdot 8\equiv (-6)^9\cdot 8 \equiv (-6)(36)^4(8) \pmod{37} \\ -48\equiv -11\equiv 26\pmod {37} \\ \therefore \boxed{ 8^{88^{888}}\pmod{37}=28}$

Now adding Chinese Remainder Theorem to the expression in boxes, that is,

$y\equiv 0\pmod 8 \\ y\equiv 1\pmod 3 \\ y\equiv 26 \pmod {37}$

We will get,

$8^{88^{888}}\pmod{888}=\boxed{ \boxed{544}}$

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P.S. : I'm very happy to solve this question :)