Simplify: $i^0+i^1+\cdots+i^{2009}.$
The powers of $i$ cycle through $i^0 = 1,$ $i^1 = i,$ $i^2 = -1,$ and $i^3 = -i,$ and the sum of any four consecutive powers of $i$ is
\[1 + i + (-1) + (-i) = 0.\]Thus, the sum reduces to $i^{2008} + i^{2009} = \boxed{1 + i}.$