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.

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.
