Prove: the biggest that N choose k gets is at k = N/2 for N even

[AxiomOfChoice]

Is there a simple way to prove this? I tried to use induction, but you get down to some horrible fraction (letting $N = 2m$ for some $m$) in the inductive step:

$$\begin{pmatrix} 2(m+1) \\ k \end{pmatrix} = \frac{(2m)!}{k!(2m - k)!} \cdot \frac{(2m+2)(2m+1)}{(2m+2-k)(2m+1-k)}$$

The inductive hypothesis is to assume the thing on the left is biggest for $k = m$, but the second fraction gets bigger as you make $k$ bigger. So...what to do! Any comments? Thanks!

 

[R136a1]

Try to prove this: If $k<k^\prime\leq N/2$, then

$$\binom{N}{k} < \binom{N}{k^\prime}$$

And if $N/2\leq k^\prime<k$, then

$$\binom{N}{k} < \binom{N}{k^\prime}$$

To prove these two statements, it suffices to look at $k$ and $k+1$.

 

[Erland]

Note that it suffices to show that k!(2m-k)! attains its minimum at k=m (0≤k≤2m), and this is equivalent to showing that
(m-r)!(m+r)! attains its minimum at r=0 (0≤r≤m).

 

[BTP]

Is there a simple way to prove this?

let n=even, k<=n/2
$$\begin{pmatrix} n \\ k \end{pmatrix} =\begin{pmatrix} n \\ k-1 \end{pmatrix} *{ (n-k+1)/(k)}$$

then (n-k+1)/k > (n/2 +1)/k >1

QED

...the symmetry between k and n-k finishes the proof.