Unit of Power of Power

$2019^{2020^{2021}} \equiv (2020-1)^{2020^{2021}} \equiv (-1)^{2020^{2021}} \equiv \boxed 1 \text{ (mod 10)}$

$9^2 \equiv 1 \mod 10$

And since $1^k \equiv 1 \mod 10$ for any integral value of k, $9^{2k} \equiv 1 \mod 10$ for any value of $k$ .

Finally, $2020^{2021}$ is an even number.

I know it is not a rigorous approach, but I don't know more than this.

Any power of a number ending in $9$ ends in $1$ or $9$ . Since only $1$ is in options, it is the answer.

2020^2021, the unit digit of the power will always be 0 as 2020 ends with 0. As we need to find out the unit digit, we can forget the first 3 digits as they will not make a change to the unit digit. Therefore we need to find out 9^2020^2021. When we keep multiplying 9 by itself, the unit digit alternates between 9 and 1. Example:9,81,729,6561, etc. Notice the unit digit alternates between 9 and 1. Therefore every odd power of 9 has unit digit 9 and every even power has unit digit 1. As we find 2020^2021 ends with 0, and zero is even, the unit digit of 2019^2020^2021 is 0