Remainder when 64 ^ 32 ^ 32 is divided by 9 is equal to _____.

Given,

${64}^{{32}^{32}} = {\left(63 + 1\right)}^{32 \times 32 \times 32 . . . . 32\text{times}}$

$⇒ {64}^{{32}^{32}} = {\left(63 + 1\right)}^{32\lambda}$

Now we know that by binomial theorem,

${\left(1 + 63\right)}^{32\lambda} = C0 {}^{32\lambda} \cdot 1 + C1 {}^{32\lambda} \cdot 1 \cdot 63 + C2 {}^{32\lambda} \cdot 1 \cdot {63}^{2} + . . . . . . .$

$⇒ {\left(1 + 63\right)}^{32\lambda} = 1 + 9\alpha$

So, when ${64}^{{32}^{32}}$ is divided by $9$ we get $1$ as remainder.