, 2003

Solution:
We must find the remainder when $2003^{2002^{2001}}$ is divided by $1000$, which will be the same as the remainder when $3^{2002^{2001}}$ is divided by $1000$, since $2003 \equiv 3 \pmod{1000}$. To do this we will first find a positive integer $n$ such that $3^{n} \equiv 1 \pmod{1000}$ and then try to express $2002^{2001}$ in the form $n k + r$, so that
$$2003^{2002^{2001}} \equiv 3^{n k + r} \equiv (3^{n})^{k} \cdot 3^{r} \equiv 1^{k} \cdot 3^{r} \equiv 3^{r} \pmod{1000}$$
Since $3^{2} = 10 - 1$, we can evaluate $3^{2m}$ using the binomial theorem:
$$3^{2m} = (10 - 1)^{m} = (-1)^{m} + 10 m (-1)^{m-1} + 100 \frac{m(m-1)}{2} (-1)^{m-2} + \cdots + 10^{m}.$$
After the first 3 terms of this expansion, all remaining terms are divisible by $1000$, so letting $m = 2q$, we have that
$$3^{4q} \equiv 1 - 20q + 100q(2q-1) \pmod{1000}$$
Using this, we can check that $3^{100} \equiv 1 \pmod{1000}$ and now we wish to find the remainder when $2002^{2001}$ is divided by $100$.
Now $2002^{2001} \equiv 2^{2001} \pmod{100} \equiv 4 \cdot 2^{1999} \pmod{4 \cdot 25}$, so we'll investigate powers of $2$ modulo $25$. Noting that $2^{10} = 1024 \equiv -1 \pmod{25}$, we have
$$2^{1999} = (2^{10})^{199} \cdot 2^{9} \equiv (-1)^{199} \cdot 512 \equiv -12 \equiv 13 \pmod{25}.$$
Thus $2^{2001} \equiv 4 \cdot 13 = 52 \pmod{100}$. Therefore $2002^{2001}$ can be written in the form $100k + 52$ for some integer $k$, so
$$2003^{2002^{2001}} \equiv 3^{52} \pmod{1000} \equiv 1 - 20 \cdot 13 + 1300 \cdot 25 \equiv 241 \pmod{1000}$$
using equation (1). So the last 3 digits of $2003^{2002^{2001}}$ are $241$.