Tower of Sevens

Since $7^{4} = 2401 \equiv 1 \pmod{100}$ , we just need to find $A = 7^{7^{7^{7^{7^{7}}}}} \pmod{4}$ .

Now $7 \equiv -1 \pmod{4}$ , so as $7^{7^{7^{7^{7}}}}$ is odd we have that $A \equiv -1 \pmod{4} \equiv 3 \pmod{4}$ .

So finally $7^{A} \pmod{100} \equiv 7^{3} \pmod{100} \equiv \boxed{43} \pmod{100}$ .

Let the number given be $N$ . We need to find $N \bmod 100$ . Since 7 is a prime, we can use Euler's theorem and Carmichael lambda repeatedly.

$\large \begin{aligned} N & \equiv 7^{7^{7^{7^{7^{7^7}}\bmod 2}\bmod 4}\bmod 20} \text{ (mod 100)} \\ & \equiv 7^{7^{7^1\bmod 4}\bmod 20} \text{ (mod 100)} \\ & \equiv 7^{7^3 \bmod 20} \text{ (mod 100)} \\ & \equiv 7^{49\times 7 \bmod 20} \text{ (mod 100)} \\ & \equiv 7^{9\times 7 \bmod 20} \text{ (mod 100)} \\ & \equiv 7^3 \text{ (mod 100)} \\ & \equiv 49\times 7 \text{ (mod 100)} \\ & \equiv (50-1)(6+1) \text{ (mod 100)} \\ & \equiv 50-6-1 \text{ (mod 100)} \\ & \equiv \boxed{43} \text{ (mod 100)} \end{aligned}$