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 noting that the last digit of the fourth powers of $1$, $3$, $7$ and $9$ is $1$ and that the last digit of every prime number $p> 5$ is $1$, $3$, $7$ or $9$.
More formally, $p=i\pmod{10}$ with $i$ in $\{1,3,7,9\}$ hence $p^4=i^4\pmod{10}$. Since every $i$ in $\{1,3,7,9\}$ is such that $i^4=1\pmod{10}$, this yields $p^4=1\pmod{10}$.
