D = d x q + r

Note $32\equiv 4\pmod{7}$ and $4^3\equiv 1\pmod{7}$ .

Moreover, $32^{32}\equiv (-1)^{32}=1\pmod{3}$ , so for some integer $n$ , $32^{32}=3n+1$ .

Then

$32^{32^{32}}\equiv 4^{3n+1}=(4^3)^n\cdot 4\equiv 1\cdot 4=\boxed{4}\pmod{7}$ .