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$\def\legendre{\genfrac(){.5pt}{}}$Let $m,n$ be positive integers. Prove the following -

  • $m!(n!)^m \mid (mn)!$
  • $m!n!(m+n)! \mid (2m)!(2n)!$

First one is here - Inductive proof
I can give a combinatorial proof. $\frac{(mn)!}{m!(n!)^m}$ is the way to split $mn$ things in $m$ groups of $n$ things.
I am looking for algebric proof of both statements.
My Work
Let $p$ be any prime
Then it suffices to prove $\legendre{m}{p} + m\legendre{n}{p} \leq \legendre{mn}{p}$ and $\legendre{m}{p} + \legendre{n}{p} + \legendre{m+n}{p} \leq \legendre{2m}{p} + \legendre{2n}{p} $ But don't know how to prove these.
It is easy to see if $p \gt min(n,m)$ then it comes obvious. But how to prove when $p \leq min(n,m)$
I'd also like to know if their exists any combinatorial prove of the second statement.
P.S: Don't give only hints. I need a full solution. I am messing up things.

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    Why is the tag (legendre-symbol) or (linear-algebra) necessary? – 2017-02-04
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    @S.C.B. I think we need legendre-symbol while proving it algebrically. – 2017-02-04
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    I think that you are going to need the floor function, not the legendre-symbol, as quadratic residues don't seem very helpful here. – 2017-02-04
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    As far as I have seen, there are no known combinatorial proofs of the second result. The only proof I have seen is based on the inequality $$\lfloor 2x\rfloor+\lfloor 2y\rfloor \geq \lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor$$ for all $x,y\geq 0$. The first result, on the other hand, has an easy combinatorial proof. – 2017-02-04
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    @S.C.B. I didn't know that there is a floor-function tag :| – 2017-02-04

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One can see that the Bivariate Catalan Number is defined by

$$C(m,n) := \frac{ (2m)!(2n)!}{m!n!(m+n)!} = \frac{\binom{2m}{m} \binom{2n}{n}}{\binom{m+n}{n}}$$

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    Choosing $m$ objects from $2m$ and $n$ objects from $2n$, I think that should be $\binom{2m}{m}\binom{2n}{n}$. Can you explain why the divide by $\binom{m+n}{n}$? – 2017-02-04
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    the Bivariate Catalan Number is an integer always so ... – 2017-02-04
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    I was talking about the previous edit – 2017-02-04
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    Why you said Choosing $m$ objects from $2m$ and $n$ objects from $2n$ ? – 2017-02-04