For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$. I am trying to prove the following statement: 

For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$. 

So far I have figured out that $n^4 = 8m$ or $n^4 = 8m + 1$.
Any help or hints are appreciated! Thanks!

An easier way is by using the powerful tool "$mod\ 8$"  for any $ n \in \mathbb{N}: n \equiv 0,1, 2, 3 ,4 , 5 , 6 \ or\   7 \ mod\ 8 \equiv 0, 1, 2, 3, 4, -3, -2 \ or\    -1 \ mod \ 8$ 
so for ${n^4}$ we have
${n^4}={n^2}^2 = n^3 n \equiv 0^4, 1^4, 2^3 \times 2 , {3^2}^2, {4^2}^2, {-3^2}^2, -2^3 \times 2 \ or\    -1^4 \ mod \ 8 \equiv 0, 1, 0, 1,0,1,0,1 \ mod\ 8 $   
I hope I didn't make it more complicated .
