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?

Alternatively:
$$\left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right)=(20^5-1)(20^{10}+20^5+1)\\
20^5\equiv (-2)^5\equiv -32\equiv 1 \pmod{11}\\
20^5\equiv 81^5\equiv 3^{20}\equiv (3^5)^4 \equiv1 \pmod{61}\\
20^5 \equiv (2^{5})^2\cdot (5^2)^2\cdot 5\equiv 1^2\cdot (-6)^2\cdot 5\equiv 25\pmod{31}\\
20^{10}\equiv (20^5)^2\equiv 25^2 \equiv (-6)^2\equiv 5\pmod{31}\\
20^{10}+20^5+1\equiv 5+25+1\equiv 0\pmod{31}$$
Note: The idea is to use the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ and decrease the exponent.
