Find the last three digits of the number $2003^{2002^{2001}}$.
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 $2003 \equiv 3(\bmod 1000)$. To do this we will first find a positive integer $n$ such that $3^{n} \equiv 1(\bmod 1000)$ and then try to express $2002^{2001}$ in the form $n k+r$, so that

$$
2003^{2002^{2001}} \equiv 3^{n k+r} \equiv\left(3^{n}\right)^{k} \cdot 3^{r} \equiv 1^{k} \cdot 3^{r} \equiv 3^{r}(\bmod 1000)
$$

Since $3^{2}=10-1$, we can evaluate $3^{2 m}$ using the binomial theorem:

$$
3^{2 m}=(10-1)^{m}=(-1)^{m}+10 m(-1)^{m-1}+100 \frac{m(m-1)}{2}(-1)^{m-2}+\cdots+10^{m} .
$$

After the first 3 terms of this expansion, all remaining terms are divisible by 1000, so letting $m=2 q$, we have that

$$
3^{4 q} \equiv 1-20 q+100 q(2 q-1)(\bmod 1000)
$$

Using this, we can check that $3^{100} \equiv 1(\bmod 1000)$ and now we wish to find the remainder when $2002^{2001}$ is divided by 100.

Now $2002^{2001} \equiv 2^{2001}(\bmod 100) \equiv 4 \cdot 2^{1999}(\bmod 4 \cdot 25)$, so we'll investigate powers of 2 modulo 25. Noting that $2^{10}=1024 \equiv-1(\bmod 25)$, we have

$$
2^{1999}=\left(2^{10}\right)^{199} \cdot 2^{9} \equiv(-1)^{199} \cdot 512 \equiv-12 \equiv 13(\bmod 25) .
$$

Thus $2^{2001} \equiv 4 \cdot 13=52(\bmod 100)$. Therefore $2002^{2001}$ can be written in the form $100 k+52$ for some integer $k$, so

$$
2003^{2002^{2001}} \equiv 3^{52}(\bmod 1000) \equiv 1-20 \cdot 13+1300 \cdot 25 \equiv 241(\bmod 1000)
$$

using equation (1). So the last 3 digits of $2003^{2002^{2001}}$ are 241.