Mods

Let $N=7^{6^n}$ , where $n=5^{4^{3^{2^1}}}$ . We need to find $N \bmod 100$ . Then, we have:

$\begin{aligned} N & \equiv 7^{6^n} \equiv 7^{2^n 3^n} \equiv (49)^{2^{n-1}3^n} \equiv (50-1)^{2^{n-1}3^n} \equiv (-1)^{\color{#3D99F6}2^{n-1}3^n} = \boxed{1} \text{ (mod 100)} \end{aligned}$ . Note that $\color{#3D99F6}2^{n-1}3^n$ is even.