The remainder when (2021) ^ 2023 is divided by 7 is

The remainder when $2021$ divided by $7$ is $- 2$. Hence, the problem reduces to finding the remainder when ${\left(- 2\right)}^{2023}$ is divided by $7$.

$= \dfrac{{\left(- 2\right)}^{2022} \times \left(- 2\right)}{7}$

$= - \dfrac{2 \times {2}^{2022}}{7}$

$= \dfrac{- 2 \times {\left({2}^{3}\right)}^{674}}{7}$

$= \dfrac{- 2 \times {\left(8\right)}^{674}}{7}$

$= \dfrac{- 2 \times {\left(1 + 7\right)}^{674}}{7}$

Using binomial theorem, we get

$= \dfrac{- 2 \times \left(1 + 7k\right)}{7}$ where $k$ is an integer.

$= \dfrac{- 2 - 14k}{7}$

Clearly, the remainder when $- 14k$ is divisible by $7$ is $0$.

$= \dfrac{- 2}{7}$

$= \dfrac{- 2 - 5 + 5}{7} = \dfrac{5}{7}$

Hence, the remainder is $5$.