Find last digit of $7^{7^{7^7}}$ Find last digit of $7^{7^{7^7}}$
I know that the last digit of $7^x$ depends on the remainder $x$ leaves when divided by $4$: ($x = 7^{7^7}$)
$$7^{4k} \equiv 1 \bmod 10$$
$$7^{4k+1} \equiv 7 \bmod 10$$
$$7^{4k+2} \equiv 9 \bmod 10$$
$$7^{4k+3} \equiv 3 \bmod 10$$
And also that 
$$7 \equiv -1 \bmod 4$$
So for all odd positive numbers, $7^{n} \equiv -1 \bmod 4$ and $7^{7}$ is odd. But which congruence relation should  I use? 
$4k+3$ and $4k+1$ are both odd 
Please explain, thanks.

(a)...$7^2\equiv 1$ mod $4\implies$ any odd power of $7$ is congruent to $7$ mod $ 4 \implies$ any odd power of $7$ is congruent to $3$ mod $4\implies 7^{7^7}\equiv 3$ mod $4$....(b)...$7^4\equiv 1$ mod $10\implies ( 7^n\equiv 7^3\equiv 3$ mod $10$ when $n\equiv 3$ mod $4)$...(c) Therefore $$7^{7^{7^7}}\equiv 3 \text { mod }10.$$
