Problem: a) Let $C_r = C_r(0)$ be the unit circle centered at origin, of radius $r>0$. Show that for $n, k \in \mathbb{N}$ with $n \geq k$ we have $$ {n\choose k} = \frac{1}{2 \pi i} \oint_{C_r} \frac{(1+z)^n}{z^{k+1}} dz. $$ b) Deduce from this that $$ {2n\choose n} \leq 4^n. $$ c) Determine the smallest value of $p > 0$ such that the inequality $$ {3n\choose n} \leq p^n$$ holds for all $n$.
My solution: a) For this part, I used a corollary Cauchy's theorem, i.e. I used $$ f^{(n)} (z) = \frac{n!}{2 \pi i} \int_C \frac{f(\eta)}{ (\eta - z)^{n+1}} d \eta. $$ In this case, we have the function $f(\eta) = (1+ \eta)^n.$ It follows that $$ \frac{f^{(k)}(0)}{k!} = \frac{1}{2 \pi i} \int_{C_r} \frac{ (1+ \eta)^n}{ \eta^{k+1}} d\eta. $$ But $f^{(k)}(0) = \frac{n!}{(n-k)!}$ so the result follows.
b) For this part, I was trying to bound the integral. From (a) we have $$ {2n\choose n}= \frac{1}{2\pi i} \oint_{C_r} \frac{(1+z)^{2n}}{ z^{n+1}} dz. $$ Now, we have $$ |\frac{1}{2\pi i} \oint_{C_r} \frac{(1+z)^2n}{ z^{n+1}} dz | \leq \frac{1}{2 \pi} \text{length}(C_r) \sup_{z \in C_r} | \frac{(1+z)^{2n}}{z^{n+1}} |. $$ I know that length of $C_r$ is $2 \pi r$. But how can I bound the other term, with the sup, such that I get $4^n$ somehow?
I think (c) can be solved if I know the approach to (b).
Thank you in advance for any help.