In the stack $7^7$⁷ the exponent is itself $7^7$, so the number is 7 raised to the power 823543. What are its last two digits?

The last two digits of the powers of 7 run 07, 49, 43, 01 and then repeat every 4 steps, so all that matters is 823543 reduced modulo 4. Since 7 leaves remainder 3 on division by 4, and an odd power of 3 leaves 3 again, 823543 ≡ 3 (mod 4) — the third entry of the cycle, 43. Checking, $7^3=343$ does end in 43. The trap 07 comes from reading the stack bottom-up as $(7^7{)}^7=7^{49}$, and 49 leaves remainder 1, landing on the first entry instead. Answering 49 stops at $7^2$ and never uses the exponent at all.