Prove ${m\choose n} \leq \frac{m^m n^{-n}}{(m-n)^{m-n}}$ using complex analysis

Question

I already know, using the Cauchy Integral Formula, that $${ \frac{1}{2\pi i} \int _{\gamma}{ \frac{z^m}{(z-1)^{n+1}}\,dz} = {m \choose n}}$$ where $\gamma$ is a positively oriented circular path around $z=1$ with arbitrary radius. I am supposed to use this result. So I know that the left hand side equals $\frac{1}{n!}f^{(n)}(1)$ with $f(z)=z^m$. Should I somehow find a bound on the $n$-th order derivative of $f$? Maybe with Cauchy estimates? Any help is appreciated.