$m,n \in \mathbb{N}, m<n\\ k=0,...,m$
Proof by calculation:
$\frac{1}{m^k} \binom{m}{k} \leq \frac{1}{n^k} \binom{n}{k}$
$\frac{1}{m^k} \binom{m}{k} = \frac{m!}{m^k * k! * (m-k)!}$
$\frac{1}{n^k} \binom{n}{k} = \frac{n!}{n^k * k! * (n-k)!}$
If $k$ has its smallest value: $k=0$
$\Rightarrow \frac{m!}{m^0 * 0! * (m-0)!} = \frac{1}{1} = 1$
$\Rightarrow \frac{n!}{n^0 * 0! * (n-0)!} = \frac{1}{1} = 1$
If $k$ has its maximum value: $k=m$
$\Rightarrow \frac{m!}{m^m * m! * (m-m)!}= \frac{1}{m^m}$
$\Rightarrow \frac{n!}{n^m * m! * (n-m)!} =$
I got stuck here. I'm not sure whether going by different cases of $k$ is the right way.
Hint:
Show that $$\frac{n^k}{m^k} \leq \frac{\binom nk}{\binom mk}$$
and use that for $i=0,\ldots , m-1$ you have $$\frac nm \leq \frac{n-i}{m-i} \Leftrightarrow m \leq n$$