Sum of powers

To find the last digit of these summation, focus on the last digits of the exponents. So, it can be written as:

$\sum_{k=0}^9 0^{(1+10k)} + \sum_{k=0}^9 1^{(2+10k)} + \sum_{k=0}^9 2^{(3+10k)} + \sum_{k=0}^9 3^{(4+10k)} + \sum_{k=0}^9 4^{(5+10k)} + \sum_{k=0}^9 5^{(6+10k)} + \sum_{k=0}^9 6^{(7+10k)} + \sum_{k=0}^9 7^{(8+10k)} + \sum_{k=0}^9 8^{(9+10k)} + \sum_{k=1}^{10} 9^{(10k)}$

Because $1^{(2+10k)} \equiv 1 \pmod{10}$ , sum of last digit is $1 \times 10 \equiv 0 \pmod{10}$ . Exponents with base 0, 5, and 6 are follows this case.

For exponents with base 4, $5+10k \equiv 1 \pmod{2}$ . Then, $4^{(5+10k)} \equiv 4 \pmod{10}$ . So, sum of last digit is $4 \times 10 \equiv 0 \pmod{10}$ . Exponents with base 9 follow this case.

For exponents with base 2, if $k$ $\begin{cases} \text{odd}, & 3+10k \equiv 1 \pmod{4} & \text{and} & 2^{(3+10k)} \equiv 2 \pmod{10} \\ \text{even}, & 3+10k \equiv 3 \pmod{4} & \text{and} & 2^{(3+10k)} \equiv 8 \pmod{10} \end{cases}$

The sum of last digit is $5 \times 2 + 5 \times 8 = 5 \times (2+8) \equiv 0 \pmod{10}$ . Exponents with base 3, 7, and 8 are follows this case.

So the last digit of summation is $0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = \boxed{0}$