"""
Self Powers
Problem 48
The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
def solution():
Returns the last 10 digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
>>> solution()
'9110846700'
total = 0
for i in range(1, 1001):
total += i**i
return str(total)[-10:]
if __name__ == "__main__":
print(solution())
