Since $m+n\geq m-n\geq 0$,${m+n\choose m-n}\geq0$, $\frac{(m+n)!}{(m-n)!}\geq(2n)!\geq(2n)(2n-2)\cdots(2)=2^n\cdot n!$ And $\frac{(m+n)!}{(m-n)!}=(m+n)(m+n-1)\cdot(m-n+1)=[(m+n)(m-n+1)]\cdot[(m+1)m]$ since $(m+i)(m+1-i)=m(m+1)+i-i^2\leq m(m+1)(1\leq i\leq n)$ So $\frac{(m+n)!}{(m-n)!}\leq[m(m+1)]^n=(m^2+m)^n$

A stronger inequality is $\binom{m+n}{m-n} \geq 2^{n}\big(n!\big)^2$.

a solution using induction on $n$ : when $n=1$ the case is clear so suppose the inequality holds for $n$ we prove it for $n+1$ $2^{n+1}(n+1)!\le\frac{(m+n+1)!}{(m-n-1)!}\le(m^2+m)^{n+1}\Longleftrightarrow\\ 2^nn!(2)(n+1)\le\frac{(m+n)!}{(m-n)!}(m+n+1)(m-n)\le(m^2+m)^n(m^2+m)$ applying the induction it suffices to prove : $2(n+1)\le (m+n+1)(m-n)\le m^2+m$ which is easy after expanding and using $m\ge n+1$