Solving an equation $\pmod {13}$ Suppose:
$$1 + \frac12 +\frac13 + \dots + \frac1{23} = \frac{a}{23!}$$
I would like to find $a \pmod {13}$.
My attempt:
I'm attempting to use Wilson's theorem which states:
$$(n-1)!= -1 \pmod n$$
consider: 
$$\begin{align}
\frac{23!a}{13} & = 23\times22\times21\dots\times14\times a\times(12!) &\\
&= -23\times22\times21\dots\times14\times a &\pmod {13} && \text{By Wilson's theorem}\\
&= -10\times9\times8\times\dots\times1\times a &\pmod{13} \\
&= -10!a &\pmod{13} \\
\end{align}$$
I have no idea where to go from here or even if I'm going in the right direction.
Any and all help is appreciated.

We have $\frac{23!}{k}\equiv 0\,\text{(mod $13$)}$ for $k=1,2,\dots,23$ except when $k=13$.
Sou you are left with $$
\begin{align}
a&\equiv \frac{23!}{13}=12!\cdot 14\cdot 15\cdots 23\\
&\equiv (-1)\cdot 1\cdot 2\cdots 10\equiv (-1)\cdot 12! \cdot 11^{-1}\cdot 12^{-1}\\
&\equiv (-1)\cdot (-1)\cdot (-2)^{-1}\cdot (-1)^{-1}\equiv -6\,\text{(mod $13$)}
\end{align}
$$
because $-2\cdot 6=-12\equiv 1\,\text{(mod $13$)}$.
