Actually it is $a^{(p-1)!} \equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} \equiv 1$ for any integer $k$, as $|(\mathbb{F}_p)^{\times}| = p-1.$
[In general, let $G$ be any group. Then for every element $g \in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (\mathbb{F}_p)^{\times}$ where the operation is multiplcation.]
Show that for every prime number $p$ and every integer $m$, the number
$mp+(pā1)!m$ is divisible by $p$.
This can be proved via two theorems: Fermat's Little Theorem and Wilson's Theorem.
$\,m^p+(p-1)!m \;=\; m^p-m+(p-1)!m+m \;=\; m^p-m+m((p-1)!+1)$
By Fermat's little theorem, $\,m^p-m\,$ is an integer multiple of $p$, since $p$ is prime.
By Wilson's Theorem, $\, (p-1)!+1\,$ is an integer multiple of $\,p\,$, since $\,p\,$ is prime.
So, for some $\, k,\,l \in \mathbb{Z}\text{,}\,$ we have
$$m^p-m+m((p-1)!+1) \;=\; kp + m(lp) \;=\; p(k+ml)\text{.}$$
Thus $\,m^p+(pā1)!m\,$ is divisible by $p$.
I believe the error in your proof came from assuming that $\,m^{p-1}\equiv1\,(\text{mod}\; p)\text{.}\,$ This is only true if we assume $m$ is not divisible by $p$.
Best Answer
I don't know whether there is some really clever way, but here a brute force approach is fast enough. $$ 70\cdot69\cdots62\equiv (-1)\cdot(-2)\cdots(-9) $$ We have $$(-8)\cdot(-9)\equiv 1\\(-2)\cdot (-5)\cdot(-7)\equiv 1\\(-3)\cdot(-4)\cdot(-6)\equiv-1$$which means that $70!\equiv61!$