Just for Some Random Person

This is straight Wilson's Theorem, which states that $(p-1)! \equiv -1 \mod p$ , for prime $p$ . Since $17$ is prime, we have that $(17-1)! \equiv -1 \mod 17 \Longrightarrow 16! \equiv \boxed{16} \mod 17$ . The proof of Wilson's Theorem is fairly elementary, only requiring some knowledge of multiplicative inverses. Basically, we can show that every number $k$ has a unique multiplicative inverse $k'$ if and only if $k$ is relatively prime to the modulus. We note that $1$ and $p-1$ are multiplicative inverses of itself, and the rest pair up, so we have $(p-1)! \equiv (p-1)\cdot(p-2)\cdots3\cdot2\cdot1 \equiv (p-1)\cdot1\cdot1 \equiv -1\cdot1\cdot1 \equiv -1 \mod p$