This can be solved by taking all terms whose base ends in a given digit, one digit at a time.

There are 202 Terms whose base ends in a 1. Their powers all end in a 1, so in total the contribute a 2 to the final solution.

There are 202 Terms whose base ends in a 2. The powers of 2 end in 2, then 4, then 8, then 6, and then repeat, so the terms of interest end in 8, then 2, then 8 etc. In total these contribute 0 to the final solution.

There are 202 Terms whose base ends in a 3. The powers of 3 end in 3, then 9, then 7, then 1, and then repeat, so the terms of interest end in 1, then 9, then 1 etc. In total these contribute 0 to the final solution.

There are 202 Terms whose base ends in a 4. The powers of 4 end in 4, then 6, and then repeat, so the terms of interest all end in 4. In total these contribute 8 to the final solution.

There are 201 Terms whose base ends in a 5. The powers of 5 all end in 5. In total these contribute 5 to the final solution.

There are 201 Terms whose base ends in a 6. The powers of 6 all end in 6. In total these contribute 6 to the final solution.

There are 201 Terms whose base ends in a 7. The powers of 7 end in 7, then 9, then 3, then 1, and then repeat, so the terms of interest end in 1, then 9, then 1 etc. In total these contribute 1 to the final solution.

There are 201 Terms whose base ends in a 8. The powers of 8 end in 8, then 4, then 2, then 6, and then repeat, so the terms of interest end in 8, then 2, then 8 etc. In total these contribute 8 to the final solution.

There are 201 Terms whose base ends in a 9. The powers of 9 end in 9, then 1, and then repeat, so all the terms of interest end in 1. In total these contribute 1 to the final solution.

There are 201 Terms whose base ends in a 0. The powers of 0 all end in 0. In total these contribute 0 to the final solution.

Finally, 2 + 0 + 0 + 8 + 5 + 6 + 1 + 8 + 1 + 0 (mod 10) = 1, so that is our final answer.