Contest math application for Wilson's theorem 
$$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{23} = \frac{a}{23!}$$ Find the remainder when $a$ is divided by $13.$

I found this online and got stuck a bit. I approached the problem as such:
From the expression we get $$a=\frac{23!}{1}+\frac{23!}{2}+\dots+\frac{23!}{23!}$$
so $$a \equiv \frac{23!}{13} \pmod{13}$$
from here we have that $$a\equiv(1\cdot 2\dots11\cdot12) \cdot(1\cdot2\dots 9\cdot10) \equiv(12!) \cdot(10!) \pmod{13}.$$
And now according to Wilson's theorem $$a \equiv(-1)(10!) \pmod{13}.$$
I cannot seem to get rid of the $10!$; what should I do here?

Since $12!=10!\times 11\times12\equiv-1\bmod13$, it follows that $10!\times-2\times-1\equiv-1\bmod13$,
so $10!\times2\equiv-1\equiv12\bmod13$, so $10!\equiv6\bmod 13$.
