Let $\phi(\cdot)$ be Euler totient function. Since 2003 is co-prime to 10 and $\phi(1000)=400$, we have \[2003^{2002^{2001}}\equiv 3^{2002^{2001}\bmod 400}\pmod{1000}.\] Note that $400=16\cdot 25.$ It is clear that $2002^{2001}\equiv 0\pmod{16}.$ On the other hand, since $\phi(25)=20,$ \[2002^{2001}\equiv 2^{2001\bmod 20}\equiv 2\pmod{25}.\] Using CRT, we have $2002^{2001}\bmod 400=352.$ Therefore, \[2003^{2002^{2001}}\equiv 3^{352}\equiv 241\pmod{1000}.\]

maxal wrote: Using CRT, we have $ 2002^{2001}\bmod 400 = 352.$ Hmm... how exactly did you compute that?

It's eay to check that both are equal $ \mod 25$ and $ \mod 16$. And nothing more is required to be shown.

ZetaX wrote: It's eay to check that both are equal $ \mod 25$ and $ \mod 16$. And nothing more is required to be shown. Hmm... I need sleep urgently.

http://www.artofproblemsolving.com/community/c6h77796p445961

Use charmichael lambda function. $\lambda(1000)=100\implies a^{100}\equiv1\pmod{1000},\text{if} \gcd(a,1000)=1$ $2003^{{2002}^{2001}}\equiv 2003^{2002^{2001}\pmod{100}}\pmod{1000}$ $\equiv 3^{2^{2001}\pmod{100}}\equiv 3^{52}\equiv241 \pmod{1000}$