Why $n!$(when represented in base fourteen) can never have $488$ trailing zeroes?
I do know that the number of zeroes on $n!$ is the number of times that $b\mid n!$.
For instance, in base fourteen, $6!$ does not have a trailing zero, since $7\nmid6$, but $7!$ does have one trailing zero.
Every multiple of $7$ contributes a trailing zero(in base fourteen), but $48!$ does have $6$ trailing zeroes, while $49!$ does have $8$ trailing zeroes, and I know this since $7^{2}\mid49$.
But I’m was stucked here:
How can a factorial(when represented in base fourteen) not have $488$ trailing zeroes?
It must be because the number of trailing zeroes jumps from less than $488$ to more than $488$. Let's see if it does.
Since there are plenty of factors $2$, it all comes down to factors of $7$. You get your first factor of $7$ at $7!$, and indeed, $7!$ has $1$ trailing zero. With $14!$ you'll get your second trailing zero, etc. And, as you already noted, with $48!$ you're at $6$ trailing zeroes, but since $49!$ gives you two factors of $7$ at once, bringing you up to $8$ trailing zeroes.
Well, we'll just have to speed up: With $(7^3-1)!=342!$ you'll have $48+6=54$ trailing zeroes, but with $343!$ you get $3$ extra, so $57$ trailing zeroes for $343!$.
Hmm, still quite a ways removed, so let's level up once more: with $(7^4-1)!=2400!$ you'll have $342+48+6=396$ trailing zeroes, so $2401!$ gives you exactly $400$ trailing zeroes.
Ah, almost there! In fact, from what we obtained before, we know that $(2401+343)!=2744!$ gives $400+57=457$ trailing zeroes.
OK, so then: $(2744+3 \cdot 49)=2891!$ will have $457+3 \cdot 8=481$ trailing zeroes.
Creeping forward now: $(2891+48)!=2939!$ will get you $481+6=487$ trailing zeroes ... but $(2891+49)!=2940!$ will get you $481+8=489$ trailing zeroes
So yes, confirmed! (and I'm sure someone with better math skills could get at this a good bit more quickly :P )