The remainder when  ${16}^{{32}^{128}}$ is divided by 7 is ____

$$\begin{gathered} {32}^{128}={\left(2^5\right)}^{128}=2^{640} \\ \begin{array}{l}now{16}^{2^{640}}={\left(2^4\right)}^{2^{640}}=2^{{4.2}^{640}}=2^{2^{642}}\\ 2^{642}={(3-1)}^{642}=3n+1forsamen\in N\\ {16}^{{32}^{128}}=2^{3n+1}\end{array} \\ \begin{array}{l}{2}^1=22^2=4,2^3=82^4=162^5=322^6=64------\\ R\left(\frac{2^1}{7}\right)=2R\left(\frac{2^2}{7}\right)=4R\left(\frac{2^3}{7}\right)=1\\ R\left(\frac{2^4}{7}\right)=2R\left(\frac{2^5}{7}\right)=4R\left(\frac{2^6}{7}\right)=1\end{array} \end{gathered}$$

Reminder when  is defined by 7

$\begin{array}{l}are=2ifm=3n+1\\ =4ifm=3n+2\\ =1ifm=3n\end{array}$