What remainder does $34!$ leave when divided by $71$? What is the remainder of $34!$ when divided by $71$?
Is there an objective way of solving this?
I came across a solution which straight away starts by stating that
$69!$ mod $71$ equals $1$ and I lost it right there.

While we can certainly show that $35! \equiv \pm 1 \pmod{71}$,
deciding between these two is apparently far from elementary.
This has been dealt with on MathOverflow,
and the answers there give the following formula,
if $p > 3$ is a prime congruent to $3$ mod $4$:
$$
\left( \frac{p-1}{2} \right)!
= (-1)^{(1 + h(-p))/2}
$$
with $h(-p)$ denoting the Class number of the field $\mathbb{Q}(\sqrt{-p})$.
In this case
the class number of $\mathbb{Q}(\sqrt{-71})$ is $7$ (source: 1, 2),
so we have
$$
35! \equiv (-1)^{\left( 1 + 7 \right)/2} = 1 \pmod{71}
$$
and
$$
34! \equiv (35)^{-1} 35! = (-2)(1) \equiv \boxed{69} \pmod{71}.
$$
