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.
 A: If $n$ is prime $p>5$ then $$n\equiv \pm1,\pm 3 \pmod{10}$$ that is final digit of $n$ is $1,3,7$ or $9$,
so $$n^2\equiv \pm1 \pmod{10}$$
that is final digit of $n^2$ is $1$ or $9$,
so $$n^4\equiv 1 \pmod{10}$$
that is final digit of $n^4$ is $1$.
A: This results directly from Euler's theorem: if $n$ is any integer coprime to $10$, $$n^{\varphi(10)}=n^4\equiv 1\mod 10,$$ and precisely, if $n$ is a prime number $>5$, it is coprime to $10$.
A: 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 
$$10| (n-2)(n-1)(n+1)(n+2) \,, \mbox{ and }\\
10|5(n-1)(n+1)$$
A: On the claim I've done in my comment above: for a prime $n > 5$ we have
$$n \equiv 1 \pmod 2 \Rightarrow n^2 \equiv 1 \pmod 8 \Rightarrow n^4 \equiv 1 \pmod{16};$$
$$n^2 \equiv 1 \pmod 3 \Rightarrow  n^4 \equiv 1 \pmod 3;\quad n^4 \equiv 1 \pmod 5$$
(the last two are implied by FLT, or may be verified by hand, considering each possible residue $\bmod {3, 5}$), so $n^4 \equiv 1 \pmod{240}$, and $240 = \gcd(7^4 - 1, 11^4 - 1)$ is the greatest possible.
