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?

$$20\equiv3^2\pmod{11}$$
$20^{15}\equiv(3^2)^{15}\equiv(3^{10})^3\equiv1^3$ by Fermat's Little Theorem
$$20=2^2\cdot5,\implies20^{15}\equiv2^{30}5^{15}$$
By Fermat's Little Theorem
$$2^{30}\equiv1\pmod{31},5^3\equiv1\pmod{31}\implies5^{15}=(5^3)^5\equiv1^5$$
Again $5^3\equiv3\pmod{61},2^6\equiv3$
$$20^{15}\equiv(2^6)^5(5^3)^5\equiv3^5\cdot3^5\pmod{61}$$
Finally
$3^5=243\equiv-1\pmod{61}$
