Determine the remainder when 15! is divided by 17.
The problem is reformulated as a linear congruence.
$$ 15! \equiv x \pmod {17} $$
Using Wilson's Theorem we have
$$ \begin{align} 16! & \equiv -1 \pmod {17} \\ \\ 16 \times 15! & \equiv -1 \pmod {17} \\ \\ 16 \times 16 \times 15! & \equiv -16 \pmod {17} \\ \\ 15! & \equiv 1\end{align} $$
The penultimate line uses $16 \times 16 \equiv (-1) \times (-1) \equiv 1 \pmod {17}$.
So the remainder when 15! is divided by 17 is 1.