Can you divide?

$32^{32^{32}}\equiv 4^{32^{32}} \equiv 4^{(-1)^{32}}\equiv \boxed{4} \pmod7$ . Since $4^3=64\equiv 1 \pmod7$ , we can work mod 3 in the exponent.

First of all we need to check out how we can write $32^{32}$ . it is equal to $(33-1)^{32}$ or in the form of $3T+1$ where T is a natural number now going back to the question $32^{32^{32}}$ is equivalent to $32^{3T+1}$ = $2^{15T+5}$ = $2^{15T+3} \times 2^{2}$ = $(7+1)^{15T+1} \times 4$ this term on evaluating by binomial expansion we get several terms but the last term will have value $1$ rest all terms will have at least $1$ exponent of $7$ and will be divisible by $7$ . so our main interest lies in the last term which is not divisible by $7$ so going back to expression we get (many terms which are divisible)*4 + $1 \times 4$ which gives us a remainder of $\boxed{4}$