1. We are asked to find the last two digits of $7^{5^6}$. 2. The last two digits of a number correspond to the number modulo 100, so we want to find $7^{5^6} \bmod 100$. 3. Use Euler's theorem: since $\gcd(7,100)=1$, the totient $\phi(100) = 40$, so $7^{40} \equiv 1 \pmod{100}$. 4. Calculate the exponent modulo 40: $5^6 \bmod 40$. 5. Compute powers of 5 modulo 40: - $5^1 = 5 \bmod 40$ - $5^2 = 25 \bmod 40$ - $5^3 = 125 = 5 \bmod 40$ (since 125 - 3*40 = 125 -120 = 5) 6. The pattern repeats every 2 powers starting from $5^1$: odd powers are 5, even powers are 25 modulo 40. 7. Since 6 is even, $5^6 \equiv 25 \pmod{40}$. 8. So the exponent reduces to $25$ modulo 40, meaning $$7^{5^6} \equiv 7^{25} \pmod{100}.$$ 9. Compute $7^{25} \bmod 100$ by successive squaring or stepwise multiplication: - $7^1 = 7$ - $7^2 = 49$ - $7^4 = 49^2 = 2401 \equiv 01 \pmod{100}$ 10. Since $7^4 \equiv 1 \pmod{100}$, write $7^{25} = 7^{4\cdot6 +1} = (7^4)^6 \cdot 7^1 \equiv 1^6 \cdot 7 = 7 \pmod{100}$. 11. Therefore, the last two digits of $7^{5^6}$ are 07. **Final answer:** The last two digits of $7^{5^6}$ are **07**.