Yeah! An Easy One...

$\phi(x)$ denotes the number of positive integers less that and coprime to $x$ .

$a^{\phi(m)} \equiv 1 \pmod m$ for all $a$ coprime to $m$ . So $14^x \equiv 14^{x~\bmod~ \phi(125)} \pmod{125}$ . $\phi(125) = 100$ .

$x$ in this case is $14^{14^{14}}$ . We only care what $14^{14^{14}}$ is in modulo $100$ to solve this problem.

We can use the same principle as above to find what $14^{14}$ is in modulo $\phi(100)=40$ , and $4^{14} \equiv 16 \pmod{40}$ . To find $x \bmod 100$ , we need to thus find $14^{16} (\bmod 100)$ , which is $36$ . So $14^x$ is the same as $14^{36} (\bmod 125)$ , so we get remainder as $\boxed{86}$ .

We can also find $14^{14^{14}} (\bmod 100)$ by finding $14^{14^{14}} (\bmod 25)$ and $14^{14^{14}} (\bmod 4)$ , and then using the Chinese Remainder Theorem.