jmc

We start from $2^1$ and consider the remainder when successive powers of 2 are divided by 7. $$\begin{array}{rl}{2}^1& \text{ leaves a remainder of 2}\\ 2^2& \text{ leaves a remainder of 4}\\ 2^3& \text{ leaves a remainder of 1}\\ 2^4& \text{ leaves a remainder of 2}\\ 2^5& \text{ leaves a remainder of 4}\\ 2^6& \text{ leaves a remainder of 1}\\ & ⋮\end{array}$$ Since 2004 is divisible by 3 (the digits sum to 6, which is a multiple of 3), we find that $2^{2005}$ leaves a remainder of $\boxed{2}$ when divided by 7.