<s>[INST] Compute the remainder when $67!$ is divided by $71$. This is how far I've been able to get.
By using Wilson's Theorem:
$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$
EDIT: Here is how I proceeded using TMM's and  Carl Mummert's hints.
$$\begin{align}
&68 \equiv -3 \pmod{71}\\
and\\
&69 \equiv -2 \pmod{71}\\
\end{align}$$
So:
$$\begin{align}
67!(-3)(-2) &\equiv 1 \pmod{71} \\
67! &\equiv 6^{-1} \pmod{71} \\
\end{align}$$
By using the Euclidean algorithm:
$$71 = 6 \cdot 11 + 5$$
$$6 = 5 \cdot 1 + 1$$
$$5 = 1 \cdot 5 + 0$$
Now, going backwards:
$$\begin{align}
1 &= 6 - 5 \\
&= 6 - (71 - 6 \cdot 11) \\
&= 6 + 6(11) - 71 \\
&= 6(1 + 11) - 71(1) \\
&= 6(12) - 71(1)
\end{align}$$
Therefore, $67! \equiv 6^{-1} \equiv 12 \pmod{71}$.
 [/INST] Recall that if 
$$xy \equiv a \pmod{n}$$
then
$$x \equiv ay^{-1} \pmod{n}$$ if $y$ is invertible in $\pmod{n}$. In your case, $n$ is $71$ a prime, which guarantees that any $y \not \equiv 0 \pmod{71}$ is invertible in $\pmod{71}$. Hence,, we have
$$67! \times 68 \times 69 \equiv 1 \pmod{71} \implies 67! \equiv (69)^{-1} (68)^{-1} \pmod{71}$$
We have
\begin{align}
68 \times 47 - 71 \times 45 & = 1\\
69 \times 35 - 71 \times 34 & = 1
\end{align}
Hence,
$$(68)^{-1} \equiv 47 \pmod{71} \text{ and }(69)^{-1} \equiv 35 \pmod{71}$$
Hence,
$$67!\equiv 35 \times 47 \pmod{71} \equiv 12 \pmod{71}$$
</s>