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I'd like to show that $${n \choose k} \ge \left( \frac{n}{k} \right) ^ k$$

I understand that $\forall \delta \ge 0$, $$\frac{n}{k} \le \frac{n-\delta}{k-\delta}$$ since as $k \le n$, then $k - \delta \le n - \delta$, thus as $\delta$ grows, the fraction grows too. But I can't show it formally.

Martin Sleziak
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Giuseppe Crinò
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    If $\delta\leq k$ then $$\tfrac{n}{k}\leq\tfrac{n-\delta}{k-\delta}\ \Leftrightarrow\ n(k-\delta)\leq k(n-\delta)\ \Leftrightarrow\ -n\delta\leq-k\delta\ \Leftrightarrow\ -n\leq -k\ \Leftrightarrow\ n\geq k.$$ – Servaes Sep 17 '15 at 10:30
  • How does that work for $n = 1, k = 0$? Are you defining $\infty^0$ somehow? – John Hughes Sep 17 '15 at 11:14

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Observe that, for $1\le k\le n$, as $\dfrac nk\ge 1$, we have $1\le\dfrac nk \le\dfrac{n-1}{k-1}$, hence $$\frac nk\le\frac{n-1}{k-1}\le\frac{n-2}{k-2}\le\dots\le\frac{n-k+1}{1},$$ whence $$\binom nk=\frac nk\cdot\frac{n-1}{k-1}\cdot\frac{n-2}{k-2}\dotsm\frac{n-k+1}{1}\ge\biggl(\frac nk\biggr)^{\!k}.$$

Bernard
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