Expression: $1^{2015}+2^{2015}+3^{2015}+\cdots+2013^{2015}+2014^{2015}$, the unit digit of the calculation result is
【Analysis and Solution】
Number Theory, Congruence.
The last digit of power operations repeats every four:
1's last digit is always 1;
2's last digits are $2, 4, 8, 6$;
3's last digits are $3, 9, 7, 1$;
4's last digits are $4, 6$;
5's last digit is always 5;
6's last digit is always 6;
7's last digits are $7, 9, 3, 1$;
8's last digits are $8, 4, 2, 6$;
9's last digits are $9, 1$;
$2015 \div 4=503 \cdots \cdots 3$;
$1^{2015}+2^{2015}+3^{2015}+\cdots+2013^{2015}+2014^{2015} \equiv 1^{3}+2^{3}+3^{3}+\cdots+2013^{3}+2014^{3}$
$=(1+2+3+\cdots+2013+2014)^{2}=[(1+2014) \times 2014 \div 2]^{2}=(2015 \times 1007)^{2} \equiv 5(\bmod 10)$;
The expression: $1^{2015}+2^{2015}+3^{2015}+\cdots+2013^{2015}+2014^{2015}$, the last digit of the result is 5.