Last digits number theory. $7^{9999}$? i have looked at/practiced several methods for solving ex: $7^{9999}$. i have looked at techniques using a)modulas/congruence b) binomial theorem c) totient/congruence d) cyclicity. 
my actual desire would be a start to finish approach using totient/congruence. i have figured out how to work the individual steps but not how to combine them and in what order to be able to solve any "end digits" questions.

Note that $7^2=49\equiv 9 \mod 10$, since $9^2=81 \equiv 1 \mod 10$, we have that
$$
7^4\equiv 1 \mod 10
$$
Now notice that $4 \mid 9996$ because $4$ divides $96$ (a number is divisible by $4$ iff its last two digits are divisible by $4$). That leaves us with a $7^3$ remaining which we know that $7^3\equiv 3 \mod 10$.
So that means that 
$$
7^{9999}\equiv 7^3 \equiv 1\cdot 3=3 \mod 10
$$
This means the last digit of $7^{9999}$ is $3$.
