Prove quotient of factorials is integral If $n$ is an integer $\gt 0$, prove 
$$\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$$
is also an integer. I understand that a general approach is proving that the power of any prime factor is greater in the numerator than it is in the denominator, but I haven't been able to formulate this into a rigorous proof.

This is a gross brute force answer.
We will show that for any positive integer $D$:
$$0\leq\left\lfloor\frac{30n}{D}\right\rfloor + \left\lfloor\frac{n}{D}\right\rfloor - \left\lfloor\frac{15n}{D}\right\rfloor - \left\lfloor\frac{10n}{D}\right\rfloor - \left\lfloor\frac{6n}{D}\right\rfloor$$
This is enough to show your theorem because when $D=p^k$ is a prime power, this is the total number of multiples of $p^k$ in the numerator minus the total number of multiples of $p^k$ in the denominator.
Write $30n = Dq+r$ for some $0\leq r < D$. Then the right hand side is:
$$q + \left\lfloor\frac{q}{30}\right\rfloor - \left\lfloor\frac{q}{2}\right\rfloor - \left\lfloor\frac{q}{3}\right\rfloor - \left\lfloor\frac{q}{5}\right\rfloor$$
Writing $q=30p+s$ with $0\leq s<30$, we see this is:
$$30p +s + p - 15p -\left\lfloor\frac{s}{2}\right\rfloor - 10p - \left\lfloor\frac{s}{3}\right\rfloor - 6p- \left\lfloor\frac{s}{5}\right\rfloor\\=s-\left\lfloor\frac{s}{2}\right\rfloor  - \left\lfloor\frac{s}{3}\right\rfloor - \left\lfloor\frac{s}{5}\right\rfloor$$
If this is non-negative for $s=0,...,29$ you are done. You can brute force from there.
Note
It seems like there should be some direct proof for:
$$0\leq s-\left\lfloor\frac{s}{2}\right\rfloor  - \left\lfloor\frac{s}{3}\right\rfloor - \left\lfloor\frac{s}{5}\right\rfloor$$
for $s=0,\dots,29$. However, it is not true for $s=30$.
