Remainder when $333^{333}$ is divided by $7$

Question

Find the Remainder when $333^{333}$ is divided by $7$

I think I have to find $333^{333}\equiv r \pmod7$ where $r(\ge0)$ is the remainder but how do I get in that form

Answer 1

$$333\equiv2^2\pmod7$$

and$$333\equiv3\pmod{\phi(7)}$$

$$\implies333^{333}\equiv(2^2)^3\equiv?\pmod7$$

Answer 2

Note that by FLT

$$333^6\equiv 1 \mod 7$$

thus

$$333^{333}\equiv 333^{3}\equiv 4^{3} \mod 7$$

Answer 3

$$333^{333}=(336-3)^{333}\equiv(-3)^{333}=-27^{111}\equiv-(-1)^{111}=1.$$