show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $ 
Let $n$ be postive integer,show that
  $$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$

I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?

Let 
$$P(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64$$  
Then $P(n+11)\equiv P(n)$ mod $11$, so if $11\mid P(n)$ for some integer $n$, then $11\mid P(n)$ for some even integer $n=2m$.  But $P(2m)=64(m^6-m^5+m^4-m^3+m^2-m+1)$.  If $11\mid P(2m)$, then, since $64\not\equiv0$ mod $11$, we must have
$$m^6-m^5+m^4-m^3+m^2-m+1\equiv0\mod11$$
After checking that $m=-1$ is not a solution, since 
$$(-1)^6-(-1)^5+(-1)^4-(-1)^3+(-1)^2-(-1)+1=7\not\equiv0\mod11$$
we can multiply both sides by $m+1$ to get the equivalent condition
$$m^7+1\equiv0\mod11$$
or $m^7\equiv-1$ mod $11$.  Cubing both sides implies $m^{21}\equiv-1$ mod $11$.  After checking that $m=0$ is not a solution, we note that $m^{10}\equiv1$, which leads to $m\equiv-1$ mod $11$, which has already been ruled out.
