I'm so excited, I used a factorial!

The power of 17 in $2017!$ is:

$\begin{aligned} v_{17}(2017) &= \left\lfloor \frac{2017}{17} \right\rfloor + \left\lfloor \frac{2017}{17^2} \right\rfloor \\ &= 118 + 6 \\ &= 124 \end{aligned}$

$2017!=17^{124} \times k,$ where $k$ is an integer such that $17 \nmid k.$

Therefore, there are $\boxed{124}$ trailing zeros in the base-17 representation of $2017!$ .