A particular proof that I don't follow. In Rudin Thm.3.20, along the proof of (d), there is an inequality that says for $n > 2k$
$${n\choose {k}}p^k \ge \frac{n^k}{2^kk!}p^k $$
I simply don't get this, especially why $n$ is restricted as such.
From the looks of it we just need to show 
$${n P k} \ge \left(\frac{n}{2}\right)^k $$
no matter what I do I cannot show that it has to be true for $n > 2k$.
Can someone give me an easy-to-follow explanation ?
 A: \begin{align}
\binom{n}{k} & = \frac{(n)\ldots(n-k+1)}{k!} \\
& = \frac{n^k}{k!} \left(\frac{n}{n}\right) \ldots \left(\frac{n-k+1}{n} \right) \\
& \geq  \frac{n^k}{k!} \left(\frac{n-k+1}{n} \right)^k \\
& \geq  \frac{n^k}{k!} \left(\frac{n - n/2 +1}{n} \right)^k \\
& \geq \frac{n^k}{k!} \frac{1}{2^k}
\end{align}
A: Your inequality is equivalent to
$$n(n-1)(n-2)\cdots(n-k+1)\ge\left(\frac n2\right)^k.\tag{1}$$
The minimum factor on the LHS is $n-k+1$, which is greater than $\frac n2$ because $n>2k$. Therefore the LHS of $(1)$ is greater than the RHS.
A: Verbose proof of this bit.
$$
(1+p)^n = \sum_{i=0}^{n}\binom{n}{i}p^i \cdot 1^{n-i} = 
\sum_{i=0}^{n}\binom{n}{i}p^i > \binom{n}{k}p^k = \qquad \text{(A)} 
$$
$$
\frac{n!}{k!(n-k)!}p^k = 
\underbrace{
n(n-1) \cdots (n-k+1) 
}_\text{number of terms is k}
\frac{p^k}{k!}= \qquad \text{(B)}
$$
$$
\underbrace{
n(n-1) \cdots 
(n-k+1)}_\text{total k terms, smallest term $(n-k+1) > \frac{n}{2}$ by assumption}
\underbrace{
\frac{p^k}{k!}
}_\text{same}
> 
\left( \frac{n}{2} \right)^k 
\underbrace{
\frac{p^k}{k!}}_\text{same} 
$$
Comments. (A) is binomial expansion, the last inequality is because we picked up only one term of the sum. (B) is just a continuation.
Key inequality of the last line holds since
$$ (n-k+1) > \frac{n}{2} $$
$$  2n-2k+2 > n $$
$$  n+2 > 2k $$
is true given that we picked $n > 2k$ by assumption.
