Divisibility of prime factorial If $p>3$ is a prime number and $q$ is the next prime following $p$, is it always true that $p!$ is divisible by all $n<q$?
 A: The answer is that $n$ divides $p!$ if $p<n<q$.
Let $p<n<q$ and assume that $r$ is a prime and $r^k\mid n$. Our goal is to prove that $r^k\mid p!$. Assume the contrary. First, note that in fact, we see that $r^k=n$. Indeed: if $n>r^k$ then $n\ge 2r^k$. On the other hand, Bertrand's postulate implies that $q<2p$ and we have
$$r^k\le\frac n2<\frac q2<p$$
and then $r^k\mid p!$.
Moreover, $k>1$. Otherwise, $r^k$ would be a prime, a contradiction with the fact that $q$ is the least prime after $p$.
Now we have
$$k>\sum_{j=1}^\infty\left\lfloor\frac p{r^j}\right\rfloor=\sum_{j=1}^{k-1}\left\lfloor\frac p{r^j}\right\rfloor\ge\sum_{j=1}^{k-1}\left\lfloor\frac {r^{k-1}}{r^j}\right\rfloor=\sum_{j=0}^{k-2}r^j\ge1+2(k-2)=2k-3$$
That is, $k<3$. This implies $k=2$.
So we have now three primes $p,q,r$ such that $p<r^2<q$ such that $\lfloor p/r\rfloor<2$, that is, $p<2r$, and then $r^2<q<2p<4r$, so $r<4$. This leaves us with two options: $n=4$ and $n=9$. But $5\le p<n$ and the last prime before $9$ is $7$, and $9\mid 7!$.
A: Obviously $p!$ is divisible by any $n \leq p$. So we need only concern ourselves with $p < n < q$.
As you have stipulated that $q$ is the very next prime greater than $p$, it follows that the numbers between $p$ and $q$ are composite numbers divisible only by primes less than or equal to $p$.
So $\gcd(n, p!) \geq 2$. Here's where your question gets interesting: it might be possible that one of these $n$ is divisible by some prime $r < p$ but with a higher exponent than in $p!$.
The likeliest candidate for such a prime is 2. Since $p$ is odd, $p!$ has $$2^\frac{p - 1}{2}$$ as a divisor.
But let's not forget that multiples of 4 contribute at least twice as much as singly even numbers to 2's exponent in $p!$. So, assuming $p \equiv 1 \bmod 4$, the larger number $$2^{\frac{p - 1}{2} + \frac{p - 1}{4}}$$ is also a divisor of $p!$ (just a small tweak if $p \equiv 3 \bmod 4$ instead).
So the best chance to accumulate enough exponents of 2 would be with the very first prime greater than a given power of 2, let's say $2^m$. But no even number between $2^m$ and $2^{m + 1}$ has a higher exponent for 2 than $m$.
I guess this is the point where we must invoke Bertrand's postulate... or maybe it's the prime number theorem that we need instead? The fact $$\frac{2^m}{\log 2^m} < \frac{2^{m + 1}}{\log 2^{m + 1}}$$ by more than 2 for $m > 4$ suggests that there are at least two primes between $2^m$ and $2^{m + 1}$.
Okay, so what if instead we choose the prime right below $2^m$? But thanks to 2 and $2^{m - 1}$, $p!$ has at least $m$ for 2's exponent in its factorization.
Surely there are some gaps in the reasoning above, but I hope at least you find it illuminating.
