High power divison by 7

Relevant wiki: Euler's Theorem

$\begin{aligned} 32^{32^{32}} & \equiv (28+4)^{32^{32}} \text{ (mod 7)} \\ & \equiv 4^{\color{#3D99F6} 32^{32} \text{ mod }\phi(7)} \text{ (mod 7)} & \small \color{#3D99F6} \text{Since }\gcd(4,7) = 1 \text{, Euler's theorem applies.} \\ & \equiv 4^{\color{#3D99F6} 32^{32} \text{ mod }6} \text{ (mod 7)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (7) = 6 \\ & \equiv 4^{\color{#3D99F6} (30+2)^{32} \text{ mod }6} \text{ (mod 7)} \\ & \equiv 4^{\color{#3D99F6} 2^{32} \text{ mod }6} \text{ (mod 7)} \\ & \equiv 4^{\color{#3D99F6}4} \text{ (mod 7)} & \small \color{#3D99F6} \text{See note for } 2^{32} \text{ mod }6 = 4 \\ & \equiv 2^8 \text{ (mod 7)} \\ & \equiv (2^3)^2(2^2) \text{ (mod 7)} \\ & \equiv (7+1)^2(4) \text{ (mod 7)} \\ & \equiv \boxed{4} \text{ (mod 7)} \end{aligned}$

---

Note: Since $\gcd(2,6) \ne 1$ , we use Chinese remainder theorem as follows:

$\begin{aligned} 2^{32} & \equiv 0 \text{ (mod 2)} \\ \implies 2^{32} & \equiv 2n & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \\ 2^{32} & \equiv (2^3)^{10}(2^2) \text{ (mod 3)} \\ & \equiv (9-1)^{10}(4) \text{ (mod 3)} \\ & \equiv (-1)^{10}(1) \text{ (mod 3)} \\ & \equiv 1 \text{ (mod 3)} \\ \implies 2n & \equiv 1 \text{ (mod 3)} \\ \implies n & \equiv 2 \\ \implies 2^{32} & \equiv 2n \equiv 4 \text{ (mod 6)} \end{aligned}$

Using Euler's Theorem. $\large a^{\phi(n)}\equiv 1\space\mod\space(n)$ Where $\phi(n)$ is the total number of integers ${1,2,3...n}$ which are not divisible with $n$

Now let's try using above theorem. $n = 7$ $\phi(n) = 6$ since $7$ is prime number and numbers not divisible with $n$ are ${1,2,3,4,5,6,}$ and $7$ is excluded as it is divisible by $7$ so $\phi(n) = 6$ writing the modular equation in Euler theorem. $\large 32^{\phi(7)} \equiv 1 \space \mod\space(7)$ $\large 32^{6} \equiv 1 \space \mod\space(7)$ Which means when $32^{6}$ is divided by $7$ it remainder is $1$ . We have $32^{1024}$ whose remainder has to be calculated and $1024\equiv 1 \space \mod \space(7)$

Now $32^{1024} = 32^1 \equiv 4 \space\mod\space(7)$ Hence the remainder is $\boxed{4}$

Note : General formula to calculate $\phi(n)$ is $\phi(n) = \left[ \left(1 - \frac{1}{p_1}\right). \left(1-\frac{1}{p_2}\right)\cdots\left(1 - \frac{1}{p_k}\right)\right]n$