Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that 
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$ 

Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} \equiv 9^{5} \bmod 11 = 9^{4} 9 \bmod 11, \:\: 9^{2} \equiv 4 \bmod 11  $$
$$ \implies 9^{5} \equiv 144 \bmod 11  \implies 20^{5} \equiv 1 \bmod 11 $$
Then the proof is done.
Now I will prove:
$$ 20^{15} \equiv 1 \bmod 31 $$ 
Notice $20^{2} \equiv 28 \bmod 31$, so 
$$20 \times (20^{2})^{7} \equiv 20 \times (28)^{7} \bmod 31 \equiv 20 \times (-3)^{7} \bmod 31 \equiv -60 \times 16 \bmod 31\equiv 32 \bmod 31 $$
then the proof is done.
Also, in similar way to prove the $20^{15} \equiv 1 \bmod 61$.
Are there shorter or more efficient proof?

$20\equiv3^2\pmod{11}$ therefore $20^{15} \equiv 3^{30}\equiv 1 \bmod11$.
$20\equiv12^2\pmod{31}$ therefore $20^{15} \equiv 12^{30}\equiv 1 \bmod31$.
$20\equiv3^4\pmod{61}$ therefore $20^{15} \equiv 3^{60}\equiv 1 \bmod61$.
