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$\large \displaystyle 7^{2k+1} \equiv 3 \pmod{4} \text{ for } k \in Z$

$\large \displaystyle 7^1 \equiv 7 \pmod{100}, \, 7^2 \equiv 49 \pmod{100}, \, 7^3 \equiv 43 \pmod{100}, \, 7^4 \equiv 1 \pmod{100}$

The Cycle repeats.

$\large \displaystyle 7^{7^{7^{7}}} \equiv 3 \pmod{4}$

Since the cycle has period $4 \pmod{100}$ , We conclude that

$\large \displaystyle \color{#D61F06}{7^{7^{7^{7}}}} \equiv \color{#3D99F6}{43} \color{#20A900}{\pmod{100}}$

$\huge \displaystyle \therefore \color{#20A900}{\text{Last }2 \text{ digits}} = \color{#D61F06}{4} \color{#3D99F6}{3}$

Since 7 and 100, $\phi (100)$ , and $\phi(\phi(100))$ are coprime integers, we can apply Euler's theorem .

$\begin{aligned} \Large 7^{7^{7^7}} & \equiv {7^{7^{7^7} \text{mod } \color{#3D99F6}{40}}} \text{(mod 100)} \quad \quad \small \color{#3D99F6}{\phi (100) = 40} \\ & \equiv 7^{7^{7^7 \text{mod } \color{#3D99F6}{16}}} \text{(mod 100)} \quad \quad \small \color{#3D99F6}{\phi (40) = 16} \\ & \equiv 7^{7^{7^{2\times 3+1} \text{mod } 16}} \text{(mod 100)} \quad \quad \small \color{#3D99F6}{7^{2\times3 + 1} \equiv 7(7^2)^3 \equiv 7(49) \equiv 7(1) \text{ (mod 16)}} \\ & \equiv 7^{7^\color{#3D99F6}{7} \text{mod }40} \text{(mod 100)} \\ & \equiv 7^{7^\color{#3D99F6}{4+3} \text{mod }40} \text{(mod 100)} \quad \quad \small \color{#3D99F6}{7^4 \equiv 2401 \equiv 1\text{ (mod 40)}, 7^3 \equiv 343 \equiv 23 \text{ (mod 40)}} \\ & \equiv 7^{23} \text{(mod 100)} \\ & \equiv 7^{4\times 5+3} \text{(mod 100)} \quad \quad \small \color{#3D99F6}{7^4 \equiv 2401 \equiv 1 \text{ (mod 100)}, 7^3 \equiv 343 \equiv 43 \text{ (mod 100)}} \\ & \equiv \boxed{43} \end{aligned}$

7^7^7^7 means (7)^343.... cyclicity of 7 repeats after 4 so we will divide 343 by 4 and the remainder will be the power we"ll put on 7....... 343/7= remainder is 3. Now 7^3 is 343 So the last two digit will be 43.

we can expand 7^7 and get last two digits 43 and if we raise 43 to the power 7 then also the result is 43 so this pattern continues and we get the answer as 43 ....