Units digit of an expression

To find the last digit of an expression which is in the format of ab we have to learn the concept of cyclicity.

Digits 0, 1, 5, 6:
These digits always give the same units digit.
${0^n} = 0; \, {1^n} = 1; \, {5^n} = 5; \, {6^n} = 6$

Digits 4, 9:
For 4, If the power is odd you will get $4$ as units digit. If the power is even, you will get $6$ as units digit.
$\begin{array}{*{20}{c}}
{{4^{odd}} = 4}\\
{{4^{even}} = 6}
\end{array}$

For 9, If the power is odd you will get $9$ as units digit. If the power is even, you will get $1$ as units digit.
$\begin{array}{*{20}{c}}
{{9^{odd}} = 9}\\
{{9^{even}} = 1}
\end{array}$

Digits 2, 3, 7, 8:
Digits 2, 3, 7, 8 follows cyclicity of $4$.
For example, 21 = 2, 22 = 4, 23 = 8, 24 = 6, 25 = 2, 26 = 4 and the cycle repeats.
Also,
$6 \times \text{Even} = \text{Even}$
$\begin{array}{l}
6 \times 2 = 2\\
6 \times 4 = 4\\
6 \times 6 = 6\\
6 \times 8 = 8
\end{array}$

Firstly, we divide the power by 4 and check the remainder.
If it is a multiple of 4, we get the following units digits for the above digits.
${2^{4n}} = {\left( {{2^4}} \right)^n} = {6^n} = 6$
${3^{4n}} = {\left( {{3^4}} \right)^n} = {1^n} = 1$
${7^{4n}} = {\left( {{7^4}} \right)^n} = {1^n} = 1$
${8^{4n}} = {\left( {{8^4}} \right)^n} = {6^n} = 6$

If the remainder is $1$,
${2^{4n + 1}} = {\left( {{2^4}} \right)^n} \times {2^1} = {6^n} \times {2^1} = 2$ \(\quad \left( \because {6 \times 2 = 2} \right)\)
${3^{4n + 1}} = {\left( {{3^4}} \right)^n} \times {3^1} = {1^n} \times {3^1} = 3$
${7^{4n + 1}} = {\left( {{7^4}} \right)^n} \times {7^1} = {1^n} \times {7^1} = 7$
${8^{4n + 1}} = {\left( {{8^4}} \right)^n} \times {8^1} = {6^n} \times {8^1} = 8$

If the remainder is $2$,
${2^{4n + 2}} = {\left( {{2^4}} \right)^n} \times {2^2} = {6^n} \times {2^2} = 4$
${3^{4n + 2}} = {\left( {{3^4}} \right)^n} \times {3^2} = {1^n} \times {3^2} = 9$
${7^{4n + 2}} = {\left( {{7^4}} \right)^n} \times {7^2} = {1^n} \times {7^2} = 9$
${8^{4n + 2}} = {\left( {{8^4}} \right)^n} \times {8^2} = {6^n} \times {8^2} = 4$

If the remainder is $3$,
${2^{4n + 3}} = {\left( {{2^4}} \right)^n} \times {2^3} = {6^n} \times {2^3} = 8$
${3^{4n + 3}} = {\left( {{3^4}} \right)^n} \times {3^3} = {1^n} \times {3^3} = 7$
${7^{4n + 3}} = {\left( {{7^4}} \right)^n} \times {7^3} = {1^n} \times {7^3} = 3$
${8^{4n + 3}} = {\left( {{8^4}} \right)^n} \times {8^3} = {6^n} \times {8^3} = 2$

Let us put all the results in a table for quick reference:

$\begin{array}{|c|c|c|c|}
\hline
Base \downarrow \\ Power \to
& 4n & 4n+1 & 4n+2 & 4n+3\\ \hline
2 & {\left( {{2^4}} \right)^n} = 6 & {\left( {{2^4}} \right)^n} \times {2^1} = 2 & {\left( {{2^4}} \right)^n} \times {2^2} = 4 & {\left( {{2^4}} \right)^n} \times {2^3} = 8\\ \hline
3 & {\left( {{3^4}} \right)^n} = 1 & {\left( {{3^4}} \right)^n} \times {3^1} = 3& {\left( {{3^4}} \right)^n} \times {3^2} = 9 & {\left( {{3^4}} \right)^n} \times {3^3} = 7\\ \hline
7 & {\left( {{7^4}} \right)^n} = 1 & {\left( {{7^4}} \right)^n} \times {7^1} = 7& {\left( {{7^4}} \right)^n} \times {7^2} = 9 & {\left( {{7^4}} \right)^n} \times {7^3} = 3\\ \hline
8 & {\left( {{8^4}} \right)^n} = 8 & {\left( {{8^4}} \right)^n} \times {8^1} = 8& {\left( {{8^4}} \right)^n} \times {8^2} = 4 & {\left( {{8^4}} \right)^n} \times {8^3} = 2\\ \hline
\end{array}$

Shortcut: For numbers 2, 3, 7, 8, find the remainder when the power is divided by 4.
If the remainder is 0, then take power as $4$.
For example, ${2^{100}}$.  The power is a multiple of 4. So ${2^{100}} = {2^4} = 6$

If the remainder is $1,\, 2,\,3$  then take the remainder as power.
For example, ${7^{50}} = {7^2} = 9$