show that if $m,n\in\mathbb{N}$ are such that $(m,n)=1$, then $$\frac{(m+n-1)!}{m!n!}\in\mathbb{N}.$$
I have a theorem (shown in my text) that says $$\fbox{If $a_1,...,a_m\in\mathbb{N}^*$ then $\frac{(a_1+...+a_m)!}{a_1!\cdot\cdot\cdot a_m!}\in\mathbb{N}$}$$ So I have $$m!n!\mid (m+n)!$$ As $(m+n)!=(m+n)(m+n-1)!$, actually only have to show that $(m!n!,m+n)=1$. And here is the question, how do I show this greatest common divisor and show that $$m!n!\nmid (m+n)$$ ...
Note that $$ \frac{(m+n-1)!}{m!n!}=\frac{m+n-1\choose m}n=\frac{n+m-1\choose n}m,$$ so if $ \frac{(m+n-1)!}{m!n!}=\frac ab$ in shortest terms, then $b\mid n$ and $b\mid m$. As $\gcd(n.m)=1$, we conclude $b=1$