please show me how to prove the following.
Given $m >= n,n\geq2$ prove
$\binom mn$ $\cdot \frac{1}{n^m} < 1$
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Given the inequalities:
$(\frac{m}{n})^n \le \binom mn \le \frac{m^n}{n!} \le (\frac{em}{n})^n$
Will that make things easier? This is the direction I take, but I still don't see any direct result by applying this inequality.
I suggest you write $m=n+a$ (so $a\geq 0$) and try to prove it by induction (on $a$).
For the induction step it may be useful that $$ \binom{p}{q}+\binom{p+1}{q}=\binom{p+1}{q+1} $$
from Pascal's triangle.
Also $p!<p^p$