Good bye $2017$

Let $N = 7^{17^{17^{2017}}}$ . We need to find $N \bmod 1000$ . Since 7, 17 and 2017 are primes, we can use Euler's theorem and the respective Carmichael lambda values are $\lambda (1000) = 100$ , $\lambda (100) = 20$ and $\lambda (20) = 4$ . Then, we have:

$\large \begin{aligned} N & \equiv 7^{17^{17^{2017 \bmod 4} \bmod 20} \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{17^{17^1 \bmod 20} \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{17^{17} \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{17^{17} \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{(20-3)^{15}(17^2) \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{-3^{15}(89) \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{3(10-1)^7(11) \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{3(69)(11) \bmod 100} \text{ (mod 1000)} \\ & \equiv 7^{77} \text{ (mod 1000)} & \small \color{#3D99F6} \text{By modular inverse (see note),} \\ & \equiv \boxed{207} \text{ (mod 1000)} \end{aligned}$

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Note:

$\begin{aligned} 7^{80} & \equiv (50-1)^{40} \equiv 1 \text{ (mod 1000)} \\ \implies 7^3 7^{77} & \equiv 1 \text{ (mod 1000)} \\ 343(7^{77}) & \equiv 1 \text{ (mod 1000)} \\ 343(7) & \equiv 401 \text{ (mod 1000)} \\ 343(207) & \equiv 1 \text{ (mod 1000)} \\ \implies 7^{77} & \equiv 207 \text{ (mod 1000)} \end{aligned}$