If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$. If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$ (ex.: $7^4-1=2400$).
How do I prove this? 

By Fermat theorem
$$p^4\equiv1\pmod{5}\\p^4-1\equiv0\pmod{5}\\p^4\equiv1\pmod{2}\\p^4-1\equiv0\pmod{2}\\p^4-1\equiv0\pmod{10}$$
Since $p$ is odd so is $p^4$  clearly it gives $1$ as remainder by $2$.Also since $2$ and $5$ are co-prime we have that $0$ is remainder of $2\cdot 5$ which gives that the last digit is $0$ 
