Prove that: $ \big(m+n\big)!\over m!n!$ is a natural number? Prove that:
(I can't prove without binomial coefficient.)

$$ \big(m+n\big)!\over m!n!$$
   is a natural number

you can use : formula for the largest power of a prime dividing a factorial
$$ n!=p_1^\alpha* p_2^\beta* ...   $$
$$ \alpha= \lfloor n/p \rfloor + \lfloor n/p^2 \rfloor+... $$
 A: Let $p$ be a prime. Write $\nu_p(k)$ for the exponent of $p$ in the prime decomposition of $k$. We have for $i \in \mathbb N$: 
$$ \frac{n+m}{p^i} = \frac n{p^i} + \frac{m}{p^i} $$
so, making the right hand side possibly smaller
$$ \frac{n+m}{p^i} \ge \left\lfloor\frac n{p^i}\right\rfloor + \left\lfloor\frac m{p^i}\right\rfloor $$
Now the right hand side is an integer, hence by definition of the floor function
$$ \left\lfloor \frac{n+m}{p^i}\right\rfloor \ge \left\lfloor\frac n{p^i}\right\rfloor + \left\lfloor\frac m{p^i}\right\rfloor $$
This gives 
\begin{align*}
 \nu_p(n!m!) &= \nu_p(n!) + \nu_p(m!)\\
     &= \sum_{i=1}^\infty \left\lfloor \frac{n}{p^i} \right\rfloor + \sum_{i=1}^\infty \left\lfloor\frac m{p^i}\right\rfloor\\
     &\le \sum_{i=1}^\infty  \left\lfloor \frac{n+m}{p^i}\right\rfloor \\
     &= \nu_p\bigl((n+m)!\bigr)
\end{align*}
And therefore
$$ n!m! = \prod_p p^{\nu_p(n!m!)} \biggm| \prod_p p^{\nu_p((n+m)!)} = (n+m)! $$
that is the denominator divides the numerator in $\frac{(n+m)!}{n!m!}$ and so the quotient is an integer.
