OK, for what it's worth, the answer can be derived by casting out all factors of $5$ and multiplying the resulting numbers incrementally $\bmod 100000$, which also allows us to also divide out from the running product the same power of $2$ as the power of $5$ that we just cast out.
As a way of checking any more elegant mathematical approach, therefore, the result is
As an aside, I'll share the process I used to calculate this and justify a shortcut that I used.
The basic idea here was to multiply successive numbers into a running product, excluding powers of $5$ and adjusting powers of $2$. The process for each number was:
- find the highest power of $5$ that divides the new number
- divide that out of the number
- multiply the adjusted number into the running product and divide by an equivalent power of $2$
- find the residue $\bmod 100000$
Note that this gives identical running $\bmod 100000$ values to @skyking's approach.
There is a intermittent problem with this approach, in that the running product will potentially be wrong for a couple of numbers after a large power of $5$ is divided out. Effectively, the value should be saturated with factors of $2$ from $8!$ onwards, but dividing out a suitable power of two may disturb this briefly. For example, the value for $15!$ is out in this process - $24368$ instead of the correct value, $74368$. Nevertheless, once a suitable number of factors of two in subsequent numbers are multiplied back into the product, the $\bmod 100000$ value gets back on track.
$2017$, occurring as it does directly after $2016$ = $2^5\cdot 63$, has an accurate running product by this method.
An continuously accurate running value can be produced, if necessary, by building up a "reserve" of a few powers of $2$ separately from a running product calculation, by stripping out powers of two from the successive values until the reserve is full. This is then used to compensate for powers of $5$ encountered, and multiplying these back in to the reduced running product. This correct gives the value for $15$ etc.