What is the last digit of $2003^{2003}$? 
What is the last digit of this number?
$$2003^{2003}$$

Thanks in advance.
I don't have any kind of idea how to solve this. Except that $3^3$ is $27$.

When we find "the last digit" of a number, it means we are working "mod $10$" (the same as "regular" arithmetic, but we only pay attention to the last digit). In particular, $ab$ (mod $10$) is the same as $[(a$ (mod $10))\cdot (b$ (mod $10))]$(mod $10$).
For example, the last digit of $37\cdot 22 = 814$ is the same as $7\cdot 2= 14$, namely: $4$.
It turns out that $3^4 = 81$ has a last digit of $1$, so:
$2003^{2003} = 3^{2003}$ (mod $10$) = $3^{750\cdot 4 + 3}$ (mod $10$)
$=(3^4)^{750}(3^3)$ (mod $10$) = $1^{750}(3^3)$ (mod $10$) = $3^3$ (mod $10$).
I daresay the last digit of $3^3$ is easily discoverable.
It isn't immediately apparent that for any number $a$ with last digit $k\in U = \{1,3,7,9\}$, that the last digit of $a^4$ is $1$, so that to find the last digit of $a^m$ we only need to find $k^{m'}$ (where $m'$ is the remainder of $m$ upon division by $4$) but this is a consequence of a far-reaching number-theoretical result known as Euler's Theorem. The down-side to using this, is that problems easy to state (such as "what is the last digit of (insert very large number here)") can be hard to explain the solutions of, without recourse to more mathematics than it looks like would be involved.
