jmc

Let's investigate the ones digits of successive powers of each of the integers from 0 through 9. At each step, we can throw away any digits other than the ones digits. Take 8 as an example: $8^1$ ends in 8, $8\times 8$ ends in 4, $8\times 4$ ends in $2$, $8\times 2$ ends in 6, $8\times 6$ ends in 8, and the pattern repeats from there. Therefore, the ones digits of $8^1,8^2,8^3,\dots$ are $8,4,2,6,8,4,2,6,\dots$. The results for all the digits are shown below.

$$\begin{array}{cc}n& \text{ones digit of }n,n^2,n^3,\dots \\ 0& 0,0,0,0,0,0,\dots \\ 1& 1,1,1,1,1,1,\dots \\ 2& 2,4,8,6,2,4,\dots \\ 3& 3,9,7,1,3,9,\dots \\ 4& 4,6,4,6,4,6,\dots \\ 5& 5,5,5,5,5,5,\dots \\ 6& 6,6,6,6,6,6,\dots \\ 7& 7,9,3,1,7,9,\dots \\ 8& 8,4,2,6,8,4,\dots \\ 9& 9,1,9,1,9,1,\dots \end{array}$$The lengths of the repeating blocks for these patterns are 1, 2, and 4. Therefore, for any digit $d$ and any exponent $a$ which is one more than a multiple of 4, the ones digit of $d^a$ is $d$. Also, if $n$ is a positive integer, then the ones digit of $n^a$ only depends on the ones digit of $n$. Therefore, for any positive integer $n$ and any exponent $a$ which is one more than a multiple of 4, the ones digit of $n^a$ is the ones digit of $n$. Let us write ``$\equiv$'' to mean ``has the same ones digit as.'' Since $2009$ is one more than a multiple of 4, we find $$\begin{array}{rl}{1}^{2009}+2^{2009}+\cdots +2009^{2009}& \equiv 1+2+3+\cdots 2009\\ & =\frac{2009(2010)}{2}\\ & =2009(1005)\\ & \equiv 9\cdot 5\\ & \equiv \boxed{5}.\end{array}$$