Asking a question related to greatest integer function in number theory This question was asked in a quiz for my course in Elementary number theory today and I am unable to think at the moment how to solve this question. 

Question: By using Greatest Integer Function , show that  $n!(n-1)!$ divides $(2n-2)!$.

I can think only that $n+n-1 >2n-2$ so dividing by $p^k$ , where $p$ is any prime and summing $k$ from $1$ to $\infty$ we get the opposite of what is asked and if both inclusions hold then both $n! (n-1)! = (2n-2)!$ Which is clearly not true. So, I think the question is wrong. 
Am I right? 
Can you please verify. 
 A: I don't see a straight forward way to do it via the greatest integer function, but for a quick proof:
We have $$\frac {(2n-2)!}{n!(n-1)!}=\frac 1{n}\times \binom {2n-2}{n-1}=C_{n-1}$$
So the claim follows by the usual properties of the Catalan Numbers
A: Fix $n$.  We want to show that, for all primes $p$, the power of $p$ in $\frac{(2n-2)!}{n!\,(n-1)!}$ is non-negative.
Fix $p$.  The power of $p$ in each of the three factorials can be found from the de Polignac formula, which is a formula that, for a non-negative integer $a$, gives $\mu(a)$, the highest power of $p$ dividing $a!$:
$$
\mu(a)=\sum_{k>0}\left\lfloor\frac{a}{p^k}\right\rfloor.
$$
Let $\ell$ be the largest exponent such that $p^\ell$ divides $n$.  From the formula we get
$$
\mu(n)=\mu(n-1)+\ell,
$$
which implies that the power of $p$ dividing our denominator, $n!\,(n-1)!$, is $2\mu(n-1)+\ell$.
For the numerator, it is clear that $\left\lfloor\frac{2n-2}{p^k}\right\rfloor\ge2\left\lfloor\frac{n-1}{p^k}\right\rfloor$, with equality holding precisely when the fractional part of $\frac{n-1}{p^k}$ is less than $\frac{1}{2}$.  For $1\le k\le\ell$ the expression $\frac{n}{p^k}$ is an integer, and, since $p\ge2$, the fractional part of $\frac{n-1}{p^k}=\frac{n}{p^k}-\frac{1}{p^k}$ is greater than or equal to $\frac{1}{2}$.  It follows that $\mu(2n-1)\ge2\mu(n-1)+\ell$.
Example: Let $n=5$.  Take the primes one at a time.
For $p=2$ we have $\ell=0$ and $\mu(5)=\mu(4)=2+1=3$.  But $\mu(8)=4+2+1=7>2\cdot3$.
For $p=3$ we have $\ell=0$ and $\mu(5)=\mu(4)=1$.  And $\mu(8)=2=2\cdot1$.
For $p=5$ we have $\ell=1$ and $\mu(5)=\mu(4)+1=0+1=1$.  And $\mu(8)=1=2\cdot0+1$.
For $p=7$ we have $\ell=0$ and $\mu(5)=\mu(4)=0$.  But $\mu(8)=1>2\cdot0$.
For $p\ge11$ we have $\ell=0$ and $\mu(5)=\mu(4)=\mu(8)=0$.
Finally $\frac{8!}{5!\,4!}=14=2\cdot7$.  Note that $2$ and $7$ were the two primes for which $\mu(8)$ exceeded $2\mu(4)+\ell$.
