Remainder

$32^1 = 7\lambda_1 + 4 \\ 32^2 = 7\lambda_2 + 2 \\ 32^3 = 7\lambda_3 + 1$

Now, $32^{32^{32}}$ divided by 7 $= {({(32)^3}^{10} \times 32^2 })^{32} \\ \implies32^{64} = {32^3}^{21} \times 32$

which gives remainder as $\boxed4$ .

I have used the equations above that $({32^3})^{\text{any integral power}}$ divided by 7 gives remainder 1 and $32^1$ divided by 7 gives remainder 4.