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?

Suppose $11\mid n^6-2n^5+4n^4-8n^3+16n^2-32n+64$. Then $n^7 + 2^7 = 0 (\mod 11)$ therefore $n^7 + 7 = 0 (\mod 11)$ equivalent $n^7= 4 (\mod 11)$. Using Fermat's little theorem, we get $1 = 4n^3(\mod 11) $ equivalent $3 = n^3(\mod 11)$
From $n^7= 4 (\mod 11)$ we get $9n = 4 (\mod 11) $ therefore $n=9(\mod 11) $. Now  $n^6-2n^5+4n^4-8n^3+16n^2-32n+64 (\mod 11)$ for $n=-2$ becomes $7 \cdot2^6 (\mod 11)$ which is $\ne 0 \mod 11$
