Digits 1

We have: $7^{2^{x}} \mod 10 = 1; x \geq 3$ $7^{7}=(2^{3}-1)^7 = k2^{3}-1 \Rightarrow 7^{7^{7}} \mod 10 = \frac{10*k +1}{7} \mod 10 =\boxed{3}$

The units digit of 7 goes like this: 7, 9, 3, 1, 7, 9, 3, 1.... Let the exponent $7^7 = x$ . We want to find the units digit of $7^x$ . Units digits of powers of $7$ repeat in cycles of 4. So we need to find the remainder when $x$ is divided by $4$ .
$x=7^7$ . $7$ is equivalent to $-1 mod 4 \quad -1^7 = -1 mod 4 = 3 mod 4.$ Therefore the units digit will be 3.

let the power of 7 be x

so cyclicity of 7 is 4 & we have the remainder when x is divided by 4

and remainder comes -1 which is not possible so remainder will be 4-1=3

and uni digit be 7 7 7=343

u digit=3

notice that the unit digit is recurring....