Alternating prime tower

We'll be using the following theorem: let $a,b,n \in \mathbb{N}$ with $\gcd(a,n)=1$ , then $a^b \equiv a^{b \mod \lambda(n)} \pmod{n}$ . ( $\lambda$ denotes the Carmichael function )

Let $x=13^{17^{13^{17}}}$ , we want to find $x \mod 1000$ . We have: $x \equiv 13^{17^{13^{17}} \mod \lambda(1000)} \equiv 13^{17^{13^{17}} \mod 100} \pmod {1000}$ Let $y=17^{13^{17}}$ , we have: $y \equiv 17^{13^{17} \mod \lambda(100)} \equiv 17^{13^{17} \mod 20} \pmod {100}$ Let $z=13^{17}$ , we have: $z \equiv 13^{17 \mod \lambda(20)} \equiv 13^{17 \mod 4} \equiv 13^1 \equiv 13 \pmod {20}$ Finally reverse the steps: $y \equiv 17^{13} \equiv 17^{3 \times 4 + 1} \equiv 4913^4 \times 17 \equiv 13^4 \times 17 \equiv 169^2 \times 17 \equiv 69^2 \times 17 \equiv 4761 \times 17 \equiv 61 \times 17 \equiv 37 \pmod{100}$ $x \equiv 13^{37} \equiv 13^{3 \times 12 + 1} \equiv 2197^{12} \times 13 \equiv \left(197^2\right)^6 \times 13 \equiv 38809^6 \times 13 \equiv \left(191^2\right)^3 \times 13 \equiv 36481^3 \times 13 \equiv 481^2 \times 481 \times 13 \equiv 361 \times 481 \times 13 \equiv 641 \times 13 \equiv \boxed{333} \pmod{1000}$