Last digit of Prime Tower

Let the number given by $N$ . We need to find $N \bmod 1000$ . Since all the numbers in the tower are prime numbers, we can apply Euler's theorem and Carmichael's lambda function . The required Carmichael's lambda values are $\lambda(1000)=100$ , $\lambda(100)=20$ , $\lambda(20)=4$ and $\lambda(4)=2$ .

$\Large N \equiv 311^{\color{#3D99F6}113^{\color{#D61F06}31^{\color{#3D99F6}13^{\color{#D61F06}3 \bmod 2} \bmod 4} \bmod 20} \bmod 100} \text{ (mod 1000)}$

Then, we have: $3 \equiv 1 \text{ (mod 2)}$ , $\implies 13^1 \equiv 1 \text{ (mod 4)}$ , $\implies 31^1 \equiv 11 \text{ (mod 20)}$ , $\implies 113^{11} \equiv 13^{11} \equiv (10+3)^{10}13^1 \equiv 3^{10}(13) \equiv (10-1)^5(13) \equiv 49(13) \equiv 37 \text{ (mod 100)}$ and

$\begin{aligned} N & \equiv 311^{37} \text{ (mod 1000)} \\ 311^3N & \equiv 311^{40} \text{ (mod 1000)} \\ (721)(311)N & \equiv (300+11)^{40} \text{ (mod 1000)} \\ 231N & \equiv 11^{40} \text{ (mod 1000)} \\ 231N & \equiv 11^{40} \text{ (mod 1000)} \\ 231N & \equiv 401 \text{ (mod 1000)} & \small \color{#3D99F6} \text{By modular reverse} \\ N & = \boxed{71} \text{ (mod 1000)} & \small \color{#3D99F6} \text{Note that } 231 \times 71 \equiv 401 \text{ (mod 1000)} \end{aligned}$