Three Two One... Wait, Add Two Thousand

[Note that this solution is not mine, I combined Pi Han's and George G's comments for clarity]

Because we're counting the last three digits, we are finding $2003^{2002^{2001}} \bmod 1000$

Modular arithmetic: $2003^{2002^{2001}} \bmod 1000 = (2003 \space \bmod 1000)^{2002^{2001}} \bmod 1000 = 3^{2002^{2001}} \bmod 1000$

By Euler's totient function, $a^{\phi (x) } \pmod {x} \equiv 1$ for coprime positive integers $a, x$

Because $3$ and $1000$ are coprime positive integers, that is, $\text{gcd} (3,1000) = 1$ , we can apply Euler's totient function

The next step is to find $\large 3^{2002^{2001} \space \bmod { (\phi(1000)) } } \bmod 1000$

To evaluate $\phi(1000)$ , we need to find all the prime factors of $1000$ , which are $2$ and $5$ , so $\phi(1000) = 1000 \times \left ( 1 - \frac {1}{2} \right ) \times \left ( 1 - \frac {1}{5} \right ) = 400$

Then, $\large 3^{2002^{2001} \space \bmod { 400 } } \bmod 1000$

Similarly like above, apply modular arithmetic: $2002^{2001} \space \bmod {400} = (2002 \space \bmod {400})^{2001} = 2^{2001} \space \bmod {400}$

You maybe tempted to apply the totient function again but note that $2$ and $400$ are not coprime. So we need to split the modulo into two (or more) numbers such that exactly one of them is divisible by $2$

$400 = 16 \times 25$

We would like to find $2^{2001} \space \bmod {16}$ and $2^{2001} \space \bmod{25}$ separately

For the first one, because $2^{2001} \div 2^4$ is obviously an integer, it is simply equals to $0$

For the second one, we can apply the totient function, $\large 2^{2001 \space \bmod { (\phi(25)) }} \space \bmod{25}$

Like above, $\phi(25) = 25 \times \left (1 - \frac {1}{5} \right ) = 20$

$\large 2^{2001 \space \bmod { 20 }} \space \bmod{25} = 2$

Apply Chinese Remainder Theorem, you should get $\large 2^{2001} \space \bmod {400} = 352$

So after all this simplification, we are left with $3^{352} \space \bmod {1000}$

$3^{352} = (10-1)^{176} \equiv {176\choose 2}10^2 - {176\choose 1}10^1 + 1 \equiv 241 \bmod 1000$

But why not start with binomial? Let $n = 2002^{2001}/2=1001\cdot2002^{2000}$ , then $2003^{2002^{2001}} \equiv 3^{2n} = (10-1)^n \equiv {n\choose 2}10^2 - {n\choose 1}10^1 + 1 \bmod 1000$ It's easy to show $n\equiv 0 \bmod 4$ and $n\equiv 1 \bmod 25$ , hence ${n\choose 2} \equiv 0 \bmod 10$ and ${n\choose 1} \equiv 76 \bmod 100$ .

use thotient function