For any prime number $n >5$, prove the final digit of $n^4$ is $1$ So I am struggling a bit with this question
$n$ is prime
we can ignore $2$ and $5$ as $n>5$
now if $n$ is prime
for the digits: $\{0,1,2,3,4,5,6,7,8,9\}$
$\{0,2,4,6,8\}$ can be discounted as $n$ cannot be even that
$5$ can be discounted as $n$ is not a multiple of $5$ either
therefore any prime must have the last digit $q$ such that $q\subset\{1,3,7,9\}$
For $n^4$, if $n=10p+q$
for a natural number $p$
I don't quite know how I can get to the desired result.

Hint 
$$n^4-1=(n^2-1)(n^2+1)=(n^2-1)(n^2-4+5)\\
=(n^2-1)(n^2-4)+5(n-1)(n+1)\\
=(n-2)(n-1)(n+1)(n+2)+5(n-1)(n+1)$$
Show that 
$$10| (n-2)(n-1)(n+1)(n+2) \,, \mbox{ and }\\
10|5(n-1)(n+1)$$
