Complex analysis ~ Binomial theorem

Given the identity

$ \binom {2n} {n} = \frac{1}{2\pi i} \int_{C_r}
\frac{(1+z)^{2n}}{z^{n+1}}dz,$

with $C_r$ the unit circle, prove that $\forall n \in \mathbb{N}$: $\binom {2n} {n} \leq 4 ^n$.

I've tried expanding the integral with the binomial theorem, but unfortunately, that doesn't seem to get me anywhere.

I'm really stuck, so all help would be dearly appreciated.

EDIT/ FOLLOW-UP: I was wondering, I think the identity as well as the result still holds when $C_r$ is an arbitrary chosen circle. Can anyone confirm this and if so, explain to me why?

Answer

$\def\abs#1{\left|#1\right|}$Note, that for $z \in C_r$, we have
$$ \abs{\frac{(1+z)^{2n}}{z^{n+1 }}} = \frac{\abs{1+z}^{2n}}{\abs z^{n+1}} = \abs{1+z}^{2n} \le 2^{2n} = 4^n $$
hence
\begin{align*}
\binom{2n}n &= \abs{\frac 1{2\pi i} \int_{C_r} \frac{(1+z)^{2n}}{z^{n+1}}\, dz}\\
&\le \frac 1{2\pi} \int_{C_r} \abs{ \frac{(1+z)^{2n}}{z^{n+1}}}\, \abs{dz}\\
&\le \frac 1{2\pi} 4^{n} \cdot 2\pi\\

&= 4^n.
\end{align*}