Prove that $1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$ I am in search of a correct and shortcut techniques to prove this.  
Otherwise I have calculated each remainder according to the power of $20$ to prove this :- 
$1 \equiv 1 \bmod 23$
$\Rightarrow 20 \equiv -3\bmod 23$
$\Rightarrow 20^2 \equiv (-3)^2 \equiv 9\bmod 23$
$20^{2n}\equiv (-3)^{2n}\bmod 23$
$20^{2n-1}\equiv (-3)^{2n-1}\bmod 23$.  
Therefore     
$1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$
For $\sum_{n=1}^{11} 20^{2n-1} \equiv r \bmod 23$
we have  $$-3-4+10-2+5-1-9+11+7-6-8 \equiv 0 \bmod 23$$    
For $\sum_{n=0}^{10} 20^{2n} \equiv s \bmod 23$
we have   $$1+9+12+16+6+8+3+4+13+2-5 \equiv 0\bmod 23$$  
Therefore $$\sum_{n=1}^{11} 20^{2n-1}+\sum_{n=0}^{10} 20^{2n} \equiv r+s \bmod 23$$
$$\Rightarrow 1+20+20^2+20^3+....+20^{21} \equiv 0+0 \equiv 0\bmod 23$$
If possible just show any short-cut correct way to prove this. Any help is appreciated.

HINT
Note that by geometric series
$$1+20+20^2+20^3+....+20^{21} \equiv \sum_{k=0}^{21}(-3)^k=\frac{1- (-3)^{22} }{4} \pmod{23}$$
then use FLT for $(-3)^{22} \pmod{23}$.
