Lower Bound for Binomial Coefficient

Theorem

Let $n, k \in \Z$ such that $n \ge k \ge 0$.

Then:

$\dbinom n k \ge \paren {\dfrac {\paren {n - k} e} k}^k \dfrac 1 {e k}$

where $\dbinom n k$ denotes a binomial coefficient.

$k! \le \dfrac {k^{k + 1} } {e^{k - 1} }$

so that:

$(1): \quad \dfrac 1 {k!} \ge \dfrac {e^{k - 1} } {k^{k + 1} }$

Then:

$\ds \dbinom n k$

$=$

$\ds \dfrac {n^\underline k} {k!}$

$\ds $

$\ge$

$\ds \dfrac {\paren {n - k}^k} {k!}$

$\ds $

$\ge$

$\ds \dfrac {\paren {n - k}^k e^{k - 1} } {k^{k + 1} }$

from $(1)$

$\ds $

$=$

$\ds \dfrac {\paren {n - k}^k e^k} {k^k} \dfrac 1 {e k}$

Hence the result.

$\blacksquare$