The proof of $(n+1)!(n+2)!$ divides $(2n+2)!$ for any positive integer $n$ 
Does $(n+1)!(n+2)!$ divide $(2n+2)!$ for any positive integer $n$?

I tried to prove this when I was trying to prove the fact that ${P_n}^4$ divides $P_{2n}$ where $n$ is a positive integer, where $P_{n}$ means the multiplication of all $k!$ from $1$ to $n$, in other words, $P_{n}=1!2!...n!$ 
So I tried a stronger statement with induction, since if I were to prove the statement 
"${P_n}^4$ divides $P_{2n}$ "
using induction, ${(n+1)!}^4$ does not divide $(2n+1)!(2n+2)!$ when $n=2$ 
So instead I tried proving ${P_n}^4 (n+1)$ divides $P_{2n}$ using induction (first motivated by the want of $n+1$ term in the dividend), but this too seems to be a bit hard since there will be a $n+2$ term in the bottom, but then I tried by plugging a few values from $1$ to $13$ into calculators, it was thus found out that maybe $(n+1)!(n+2)!$ divides $(2n+2)!$. 
I tried this by first considering the number of $p$ (prime) dividing the divisor must be smaller or equal to the number of the $p$ dividing the dividend. Namely:
$(\left \lfloor{\frac{n+1}{p}}\right \rfloor + \left \lfloor{\frac{n+1}{p^2}}\right \rfloor +...)$+ $(\left \lfloor{\frac{n+2}{p}}\right \rfloor + \left \lfloor{\frac{n+2}{p^2}}\right \rfloor +...)$ $\leq$ $\left \lfloor{\frac{2n+2}{p}}\right \rfloor + \left \lfloor{\frac{2n+2}{p^2}}\right \rfloor +...$.
Then by letting $j$ be an arbitrary positive integer, it was proven that 
$\left \lfloor{\frac{n+1}{p^j}}\right \rfloor + \left \lfloor{\frac{n+2}{p^j}}\right \rfloor \leq \left \lfloor{\frac{2n+2}{p^j}}\right \rfloor$
by considering many cases of whether $p^j$ divides $n+1$.
Is there any other proof more intuitive and more "elegant" than this one where we have to consider the many cases? Or is there even any better approach of proving the original problem? (preferably an attempt of both method, induction and without induction.)
Thanks
Than

Although the expression $\frac{(2n+2)!}{(n+1)!(n+2)!}$ is not quite a binomial coefficient, it can be expressed as the difference of two binomial coefficients (and so it is in fact an integer):
$$ \frac{(2n+2)!}{(n+1)!(n+2)!} = \frac{1}{n+2} \binom{2n+2}{n+1} = \binom{2n+2}{n+1} - \binom{2n+2}{n+2} $$
This restates that the Catalan number $C_{n+1}$ is a whole number.
