The condominium of numbers!

Let, $x=2^{3^{4^{5^{6^{7^{8^{9}}}}}}}$ and then we have to find $x \pmod {1000}$ .

As $1000=8×125$ , we can use Chinese Remainder Theorem as follows,

Finding $x \pmod 8$ ,

$\boxed{x= 2^{3^{4^{5^{6^{7^{8^{9}}}}}}} \equiv 0 \pmod 8}$

Finding $x \pmod{125}$ ,

$\begin{aligned} 2^{3^{4^{5^{6^{7^{8^{9}}}}}}} & \pmod {125} \\ \Rightarrow 3^{4^{5^{6^{7^{8^{9}}}}}} & \pmod {\phi(125)=100} \\ \Rightarrow 4^{5^{6^{7^{8^{9}}}}} & \pmod{\lambda(100)=20} \end{aligned}$

Now observing that,

$4^k \pmod {20} , \forall k\in \mathbb{N} \\ \begin{cases} 4^{\text{odd}} \equiv 4 \pmod{20} \\ 4^{\text{even}} \equiv 16 \pmod{20}\end{cases}$

Therefore,

$\begin{aligned} 4^{5^{6^{7^{8^{9}}}}} & \equiv 4 \pmod{\lambda(100)=20} \\ \Rightarrow 3^{4} & \equiv 81 \pmod{\phi(125)=100} \\ \Rightarrow 2^{81} & \equiv (2^7)^{11} \cdot 2^4 \pmod {125} \\ 3^{11}\cdot 16 & \equiv (3^5)^2 \cdot 3\cdot 16 \pmod{125} \\ 7^2\cdot 3 \cdot 16 & \equiv 102 \pmod{125} \end{aligned}$

Therefore,

$\boxed{x\equiv 102 \pmod{125}}$

Now applying Chinese Remainder Theorem on the expressions in the boxes,

$x\equiv \boxed{ \boxed{ 352}} \pmod{1000}$

We are given $n=2^{3^{4^a}}$ where $a$ is odd. If we work modulo 125 in the base and modulo $\lambda(\lambda(125))$ $=\lambda(100)$ $=20$ in the second exponent, $4^a$ , we observe that $4^a\equiv 4 \pmod{20}$ for odd $a$ so that $n\equiv 2^{3^4}$ modulo 125 and modulo 1000. Now $2^{78}\equiv 44 \pmod{125}$ so that $n\equiv 2^{3^4}=2^{81}\equiv 8\times 44=\boxed{352} \pmod{1000}$

I think that one of the fastest method to solve the problem is using a mix of eulero φ function and chaermical (or λ) function. So working modulo 1000 the φ is 400 while λ is 200. With some calculus we can obtain first $8^9$ and then all the other powers. At $6th$ power we have $6^{801}$ wich is $6^1$ thanks to φ. $5^6 \equiv 625 \pmod {1000}$ here I used the two functions to reduce 625 to 25. Then $4^{25} \equiv 624 \pmod {1000}$ (also here from 624 to 24). $3^{24} \equiv 481 \pmod{1000}$ and the last calculus is $2^{81}$ wich is $352 \pmod {1000}$ . So the solution is $\boxed{352}$

To be certain, I applied computing method to take least significant figures.