Show that the following relation holds for any positive integers $1 < k \le m < n$: $$\binom{n}{k}m^k > \binom{m}{k}n^k$$
Because $$\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!}$$
and $$\binom{m}{k}=\frac{m(m-1)\cdots(m-k+1)}{k!}$$
therefore the to-be-proved inequality is equivalent to $$\frac{n(n-1)\cdots(n-k+1)}{n^k} > \frac{m(m-1)\cdots(m-k+1)}{m^k}$$
This indeed holds because for any integer $i = 1, 2, \cdots, k-1$, $$\frac{n-i}{n} = 1-\frac{i}{n} > 1 - \frac{i}{m} = \frac{m-i}{m}$$
Multiplying these $(k-1)$ inequalities gives the desired result.