What's special about 11 ?

$\begin{aligned} \sum_{n=0}^{19}n^{n+1}&=-20^{21}-21^{22}+\sum_{n=0}^{21}n^{n+1}\\&\equiv-(-2)^{21}-(-1)^{22}+\sum_{n=0}^{21}n^{n+1}\\&\equiv1+\sum_{n=0}^{21}n^{n+1}\\&=1+\sum_{n=0}^{10}(n^{n+1}+(n+11)^{n+12})\\&\equiv1+\sum_{n=0}^{10}(n^{n+1}+n^{n+2})\\&\equiv1+\sum_{n=0}^{10}(n^{n+1}+(10-n)^{12-n})\\&\equiv1+\sum_{n=0}^{10}(n^{n+1}+(-1-n)^{2-n})\pmod{11} \end{aligned}$

and on and on like this, just pair up the residues

lol manual ... 38312573763282605864751634 mod 11 = 4 :v

This Solution Is For Those, Who Don't Understand the More Advanced Ones.(me included) We use the property of congruence that 12=1 mod 11 and 12^n==1^n mod 11. and also Little Fermat's theorem a^p=a mod p(if p is prime). since 11 is prime, we can use this. we change every base from 11 up to lower one 12->1; 13->2,... and power from 11 in this manner 11->1; 12->2 (since a^(11+x)=a^11 * a^x=a * a^x=a^(x+1) mod 11). Then when we have done this we just take some numbers at the time and find remainder on the calculator