An Old Number theory IMO question In my book, under Legendre’s Function, the following two examples were given;
When $m,n \in N$ , prove that;

*

*$\ $ $m! \cdot (n!)^m$ divides $(mn)!$


*$\ $ $m! \cdot n! \cdot (m+n)! $ divides $(2m)! \cdot (2n)!$
Well, for the first one, I know it is just the number of ways to put $mn$ balls into $m$ identical baskets, each with $n$ balls.
But, as this is a NT book, I tried solving this with Legendre’s Function and it just did not work,
I got we needed to prove:
$f(mn) \ge f(m) + m \cdot f(n)$ where $f(k) = [\frac{k}{p}]$ where [.] is the floor function.
Now, I could not figure out what to do, I tried inducting on $n$ but after using some inequalities like $[xy] \ge [x][y]$ and $[x+y] \ge [x]+[y]$ , the inequality just became false.
As I couldn't even solve the first one, I could not solve the second one either, not even 'combinatorically'.
So, I am looking for a proof of both the problems using Legendre’s Function (and perhaps a combinatorial proof of $2$ well?)
Thanks!

Question $\bf{1}$
In this answer, it is shown that
$$
\frac{(mn)!}{(m!)^nn!}=\prod_{k=1}^n\binom{mk-1}{m-1}\tag1
$$

Question $\bf{2}$
Using the equation $x=\lfloor x\rfloor+\{x\}$, we get the equation
$$
\lfloor 2x\rfloor+\lfloor 2y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor-\lfloor x+y\rfloor=\{x\}+\{y\}+\{x+y\}-\{2x\}-\{2y\}\tag2
$$
There are two possibilities: if $\{x\}+\{y\}\lt1$
$$
\{x+y\}=\{x\}+\{y\}\tag{3a}
$$
or if $\{x\}+\{y\}\ge1$
$$
\{x+y\}=\{x\}+\{y\}-1\tag{3b}
$$
If $\{x\}\lt\frac12$ and $\{y\}\lt\frac12$, then $\text{(3a)}$ implies
$$
\{x\}+\{y\}+\overbrace{\{x+y\}}^{\{x\}+\{y\}}-\overbrace{\{2x\}}^{2\{x\}}-\overbrace{\{2y\}}^{2\{y\}}=0\tag{4a}
$$
If $\{x\}\ge\frac12$ and $\{y\}\ge\frac12$, then $\text{(3b)}$ implies
$$
\{x\}+\{y\}+\overbrace{\{x+y\}}^{\{x\}+\{y\}-1}-\overbrace{\{2x\}}^{2\{x\}-1}-\overbrace{\{2y\}}^{2\{y\}-1}=1\tag{4b}
$$
Otherwise, assume $\{x\}\ge\frac12$ and $\{y\}\lt\frac12$, then if $\{x\}+\{y\}\,{\color{#C00}{\lt}\atop\color{#090}{\ge}}\,1$
$$
\{x\}+\{y\}+\overbrace{\{x+y\}}^{\{x\}+\{y\}-{\color{#C00}{0}\atop\color{#090}{1}}}-\overbrace{\{2x\}}^{2\{x\}-1}-\overbrace{\{2y\}}^{2\{y\}}={\color{#C00}{1}\atop\color{#090}{0}}\tag{4c}
$$
Thus, $(2)$ and $(4)$ ensure that
$$
\lfloor 2x\rfloor+\lfloor 2y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor-\lfloor x+y\rfloor\ge0\tag5
$$
Therefore, for any prime $p$,
$$
\overbrace{\sum_{k=1}^\infty\left(\left\lfloor\frac{2n}{p^k}\right\rfloor+\left\lfloor\frac{2m}{p^k}\right\rfloor\right)}^\text{factors of $p$ in $(2n)!(2m)!$}\ge\overbrace{\sum_{k=1}^\infty\left(\left\lfloor\frac{n}{p^k}\right\rfloor+\left\lfloor\frac{m}{p^k}\right\rfloor+\left\lfloor\frac{n+m}{p^k}\right\rfloor\right)}^\text{factors of $p$ in $n!m!(n+m)!$}\tag6
$$
That is,
$$
n!\,m!\,(n+m)!\mid(2n)!\,(2m)!\tag7
$$
