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?

Starting from @Crostul's comment, we have
$$
f(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64 = \frac{n^7+2^7}{n+2}
$$
and so $f(n)(n+2)=n^7+2^7$.
Therefore, if $11$ divides $f(n)$, then $11$ divides $n^7+2^7$.
Now $n \mapsto n^7$ is an injective map mod $11$ because $\gcd(7,\phi(11))=1$.
Therefore, if $11$ divides $f(n)$, then $n^7 \equiv -2^7 = (-2)^7 \bmod 11$ and so $n \equiv -2 \bmod 11$.
So you only have to check that $11$ does not divide $f(n)$ for $n=-2$.
