nicetry007 wrote: For $ q \;<\; n$, $ \;q(q-1)\; |\; (n-1)!$. Hence, it is enough discuss the problem for $ q = n$. If $ n$ is composite, then $ n = a \cdot b$ and $ gcd(n-1,a) = gcd(n-1,b) = gcd(n-1,n) = 1$. Both $ a$ and $ b$ are less than $ n-1$. Hence, $ ab(n-1)\; |\; (n-1)!$. We are left with the case of $ n$ being a prime. Let $ n = p$. We know that $ (p-1)! =-1 (mod\; p)$. In other words, $ (p-1)! = (p-1)(p-2)! = pk-1 = p(k-1)+p-1 \Rightarrow (p-1)\left[(p-2)!-1\right] = p(k-1) \Rightarrow (p-1) \;| \; k-1$. But $ \lfloor\frac{(p-1)!}{p}\rfloor = k-1$. Hence, $ (p-1) \;| \; \lfloor\frac{(p-1)!}{p}\rfloor$

for $q<n$ is obvious and for $n=q$ only note that $\left\lfloor \frac{(q-1)!}{q}\right\rfloor =\frac{(q-1)! +1}{q} -1$