Cyclical digits

Relevant wiki: Euler's Theorem

We need to find $12577^{335} \text{ mod 10}$ .

$\begin{aligned} 12577^{335} & \equiv (12570 + 7)^{335} \text{ (mod 10)} \\ & \equiv 7^{\color{#3D99F6}335} \text{ (mod 10)} & \small {\color{#3D99F6}\text{Since }\gcd(7,10) = 1 \text{, Euler's theorem applies.}} \\ & \equiv 7^{\color{#3D99F6}335 \text{ mod }\phi(10)} \text{ (mod 10)} & \small {\color{#3D99F6}\text{Euler's totient function }\phi (10) = 4} \\ & \equiv 7^{\color{#3D99F6}335 \text{ mod }4} \text{ (mod 10)} \\ & \equiv 7^{\color{#3D99F6}3} \text{ (mod 10)} \\ & \equiv 7\cdot 49 \text{ (mod 10)} \\ & \equiv 7(50-1) \text{ (mod 10)} \\ & \equiv -7 \text{ (mod 10)} \\ & \equiv \boxed{3} \text{ (mod 10)} \end{aligned}$