A Huge Summation it is. #1

Let's write all items of the summation:

${11}^{11}=(10+1)^{11}=10\times{C_1}+{1}^{11}\\{22}^{22}=(20+2)^{22}=10\times{C_2}+{2}^{22}\\{33}^{33}=(30+3)^{33}=10\times{C_3}+{3}^{33}\\\vdots\\(90+9)^{99}=10\times{C_9}+{9}^{99}$

It's clear that $10\times{C_1}, 10\times{C_2}, 10\times{C_3}, \ldots, 10\times{C_9}$ are divisible by $10$ , so we just need to find the reminder of ${1}^{11}, {2}^{22}, {3}^{33}, \ldots, {9}^{99}\pmod{10}$ .

By using Euler's theorem, CRT, and the properties of modular arithmetic we will find that

${2}^{22}\equiv4\pmod{10}~~~~~{3}^{33}\equiv3\pmod{10}~~~~~{4}^{44}\equiv6\pmod{10}\\{5}^{55}\equiv5\pmod{10}~~~~~{6}^{66}\equiv6\pmod{10}~~~~~{7}^{77}\equiv7\pmod{10}\\{8}^{88}\equiv6\pmod{10}~~~~~{9}^{99}\equiv9\pmod{10}$

Now, we have

$1+4+3+6+5+6+7+6+9\equiv47\equiv\boxed{7}\pmod{10}$