Estimating standard deviation: divide by n-1

Two reasons for that

To reduce gap between sample variance and population variance
( empirical reason )

"1/n" version is the maximum likelihood estimate of the population variance, however, it is also mathematically biased
sample variance is usually smaller than the population variance
→ estimation of the population variance is getting bigger than real
to reduce gap using "1/n-1" convention ( provides an unbiased estimate )
why not n-2 ?

To match both expectation of sample variances and population variance
( mathematical reason )

let,
$n$ : sample size
$\bar{X}$ : sample mean
$s^2$ : sample variance
$m$ : population mean
$\sigma^2$ : population variance
then, figure out following is true
$E[s^2] = 1/(n-1) E[\sum\limits_{k=1}^n (X{k}-\BAR{x})^2]$
first,
$\sum\limits_{k=1}^n (X{k}-\BAR{x})^2 = \sum\limits_{k=1}^n ((X{k}-m) + (m-\bar{X}))^2$
$\thickspace=\sum\limits_{k=1}^n ((X{k}-m)^2 + 2(X{k}-m)(m-\bar{X}) + (m-\bar{X})^2)$
$\thickspace=\sum\limits_{k=1}^n ((X{k}-m)^2 + 2(\bar{X}-m)n(m-\bar{X}) + n(m-\bar{X})^2$
$\thickspace=\sum\limits_{k=1}^n ((X{k}-m)^2 + 2(\bar{X}-m)n(\bar{X}-m) + n(\bar{X}-m)^2$
$\thickspace=\sum\limits_{k=1}^n ((X{k}-m)^2 -m(\bar{X}-m)^2$
$\therefore E[s^2] = 1/(n-1) E[\sum\limits_{k=1}^n (X{k}-\BAR{x})^2]$
$\thickspace= 1/(n-1) E[\sum\limits_{k=1}^n (X{k}-m)^2 - n(\BAR{x}-m)^2]$
as here,
$E[(X{k}-m)^2] = \sigma^2$
$E[(\bar{X}-m)^2] = V(\bar{X}) = \sigma^2/n$
$1/(n-1) E[\sum\limits_{k=1}^n (X{k}-m)^2 - n(\BAR{x}-m)^2] = 1/(n-1) * (n\sigma^2-n(\sigma^2/n)) = \sigma^2$
$\therefore s^2 = 1/(n-1) \sum\limits_{k=1}^n (X{k}-\bar{X})^2$