Funny exponent

We need to find $7^{10^6} \bmod 10^8$ . Let us consider $2^8$ and $5^8=$ using Chinese remainder theorem .

$\begin{aligned} 7^{10^6} & \equiv 49^{5\times 10^5} \text{ (mod }2^8) \\ & \equiv (48+1)^{5\times 10^5} \text{ (mod }2^8) \\ & \equiv (\cdots + 48\times 5 \times 10^5+1) \text{ (mod }2^8) \\ & \equiv 1 \text{ (mod }2^8) \end{aligned}$

$\begin{aligned} 7^{10^6} & \equiv 49^{5\times 10^5} \text{ (mod }5^8) \\ & \equiv (50-1)^{5\times 10^5} \text{ (mod }5^8) \\ & \equiv (\cdots - 50\times 5 \times 10^5+1) \text{ (mod }5^8) \\ & \equiv 1 \text{ (mod }5^8) \end{aligned}$

This means that $7^{10^6} \equiv 2^8n+1$ , where $n$ is an integer. And $2^8n +1 \equiv 1 \text{ (mod }5^8)$ $\implies n = 0$ and $7^{10^6} \equiv \boxed 1 \text{ (mod }10^8$ .

Since $7^4 = 2401$ , then $7^{10^6} = (24\times 10^2 + 1)^{10^6 /4}$ gives a remainder of $\boxed1$ when divided by $10^8$ .