Wilson's theorem and fractions 
What is the remainder of $m$ satisfying $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3} +\dots+\frac{1}{33}=\frac{m}{33!}$$ upon division by $17$?

The factorial here hints me towards Wilson's theorem, but I cannot seem to get this in any way to the form such that I could use it. How should I go about this?

Note that $$m=\frac{33!}{1}+\frac{33!}{2}+\cdots+\frac{33!}{33}$$
And we have to find the remainder when $m$ is divided by $17$.
Note that out of all the above terms, all terms except $\dfrac{33!}{17}$ are divsible by $17$. Thus $m\equiv \frac{33!}{17} \pmod{17}$.
But we have $$\begin{aligned}\frac{33!}{17}&=33\times 32\times \cdots \times 18 \times 16\times \cdots \times 1 \\&\equiv (-1)\times (-2)\times\cdots\times (-16)\times16\times\cdots \times1 \pmod{17} \\ &\equiv (16!)^2\pmod{17}\end{aligned}$$
Now from Wilson's theorem, we have $16!\equiv -1\pmod{17} \implies (16!)^2\equiv 1 \pmod{17}$
Thus $m\equiv \boxed{1} \pmod{17}$
