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I can calculate module exponentiation of square powers of $7$ modulo $35$ or things alike by hand but for computing modulo exponentiation of square powers of $7$ modulo $1000$, then I need a calculator. For instance

$$ \begin{alignat*}{2} 7^1&\equiv7&&\pmod{1000}\\ 7^2&\equiv49&&\pmod{1000}\\ 7^4&\equiv(7^2)^2\equiv49^2\equiv401&&\pmod{1000} \\ 7^8&\equiv(7^4)^2\equiv401^2\equiv801&&\pmod{1000} \\ 7^{16}&\equiv(7^8)^2\equiv801^2\equiv601&&\pmod{1000}\\ 7^{32}&\equiv(7^{16})^2\equiv601^2\equiv201&&\pmod{1000}\\ 7^{64}&\equiv(7^{32})^2\equiv201^2\equiv401&&\pmod{1000} \end{alignat*} $$ Is there an easier way without using a calculator?

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You can do it by hand, if you observe $7$ has order mod. $1000$ a divisor of $\varphi(1000)=80$. Indeed, you can check by hand that $$7^{10}=7^2\cdot 7^8\equiv 49\cdot 801\equiv 249\mod 10000, \qquad 7^{20}\equiv 7^4\cdot7^{16}\equiv 401\cdot 601\equiv 1\mod 1000.$$ Thus $\;7^{2^n}\equiv7^{2^n\bmod20}$. Then you can use that the powers of $2$ mod .$20$ are cyclic: $$\begin{array}{r|ccccccc} n&1&2&3&4&5&6&\dots\\ \hline 2^n\bmod 20&2&\color{red}4&8&16&12&\color{red}4&\dots\end{array}$$

Bernard
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