Big number problem

We can work modulo 100 in the base and modulo $\phi(100)=40$ in the exponent. $2007^{2007}\equiv 7^7 \equiv 7^3\equiv 43 \pmod{100}$ so that the answer is $\boxed{4}$

Consider the same number written in binomial form: $(2000 + 7)^{2007}$ With binomial expansion, one can note that no powers of 2000 will contribute to the first 3 digits of the number we are looking for. Thus, we can just focus on powers of 7.

Note the first few powers of 7:

1. 07
2. 49
3. 343
4. 2401
5. 16807

When paying attention to the last 2 digits we can the pattern emerge: 07, 49, 43, 01, 07 and so on.

This pattern continues infinitely. To find the 2007th entry of this sequence one can look at the number 2007mod4 (since there are only 4 terms in this pattern). 2007mod4 = 3.

Thus the 2007th term of this sequence is the same as the 3rd term of the sequence, 43. Hence the digit we look for is the number 4.