Again the same?

Note that $7^2 \equiv 1 \pmod{16}$ , which means that $7^{2^n} \equiv 1 \pmod{2^{n+3}}$ for all integers $n \ge 1$ . Similarly $7^4 \equiv 1 \pmod{25}$ , and hence $7^{4 \times 5^n} \equiv 1 \pmod{5^{n+2}}$ for all $n \ge 0$ . We certainly deduce that $7^{10^n} \equiv 1 \pmod{10^{n+2}}$ for all $n\ge 2$ . We also note that $7^{20} \equiv 1 \pmod{1000}$ .

Letting ${}^{(n)}7$ denote that level $n$ tetration of $7$ , we see that ${}^{(2)}7 = 7^7 \equiv 3 \pmod{20}$ , and hence ${}^{(3)}7 \equiv 7^3 \equiv 343 \pmod{1000}$ . Thus $\begin{array}{rclcrcl} {}^{(3)}7 & \equiv & 43 \pmod{10^2} & \Rightarrow & {}^{(4)}7 & \equiv & 7^{43} \equiv 2343 \pmod{10^4} \\ {}^{(4)}7 & \equiv & 343 \pmod{10^3} & \Rightarrow & {}^{(5)}7 & \equiv & 7^{343} \equiv 72343 \pmod{10^5} \\ {}^{(5)}7 & \equiv & 2343 \pmod{10^4} & \Rightarrow & {}^{(6)}7 & \equiv & 7^{2343} \equiv 172343 \pmod{10^6} \\ {}^{(6)}7 & \equiv & 72343 \pmod{10^5} & \Rightarrow & {}^{(7)}7 & \equiv & 7^{72343} \pmod{10^7} \end{array}$ While we have used the fact that $7^{10^5} \equiv 1 \pmod{10^7}$ , the results in the first paragraph show that $7^{50000} \equiv 1 \pmod{10^7}$ , and hence ${}^{(7)}7 \equiv 7^{22343} \pmod{10^7}$ In fact $50000$ is the order of $7$ modulo $10^7$ , and so the desired answer is $\boxed{22343}$ .