Power Tower!

Let $N = 5^{4^{3^{2^1}}}$ . We need to find $N \bmod 100000$ . Since $\gcd(5,100000) \ne 1$ we cannot use the Euler's theorem and we have to consider the factors of 100000, $2^5 = 32$ and $5^5 = 3125$ separately with the Chinese remainder theorem .

Consider factor 32:

$\begin{aligned} N & \equiv 5^{\color{#3D99F6}4^9 \bmod \phi(32)} \text{ (mod 32)} & \small \color{#3D99F6} \text{Since }\gcd(5,32)=1 \text{, Euler's theorem applies.} \\ & \equiv 5^{\color{#3D99F6}4^9 \bmod 16} \text{ (mod 32)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(32) = 16 \\ & \equiv 5^0 \text{ (mod 32)} \\ & \equiv 1 \text{ (mod 32)} \end{aligned}$

Consider factor 3125:

$\begin{aligned} N & \equiv 0 \text{ (mod 3125)} \\ \implies N & \equiv 3125n & \small \color{#3D99F6} \text{where }n \in \mathbb N \\ \implies 3125n & \equiv 1 \text{ (mod 32)} \\ 21n & \equiv 1 \text{ (mod 32)} \\ \implies n & \equiv 29 \end{aligned}$

$\implies N \equiv 3125(29) \equiv \boxed{90625} \text{ (mod 100000)}$