If  7 divides  ${32}^{{32}^{32}}$ , the remainder is 

 $$\begin{gathered} 32=2^5\Rightarrow {\left(32\right)}^{32}={\left(2^5\right)}^{32} \\ =2^{160}={(3-1)}^{160}=3m+1,m\in N \\ \therefore {\left(32\right)}^{{32}^{32}}={\left(32\right)}^{3m+1}=2^{5(3m+1)} \\ {2}^{3(5m+1)}{2}^2={4.8}^{5m+1} \\ 4{(7+1)}^{5m+1}=4(7n+1),n\in N=28n+4 \end{gathered}$$
$\therefore$   When 7 divides  ${\left(32\right)}^{{32}^{32}}$  remainder =4