When calculating confidence intervals for a population with standard deviation σ unknown, σ is estimated using the sample standard deviation S, which uses the Bessel correction to more closely approximate the real σ.
But suppose the population X is a Bernoulli variable. X being now a binary variable, $ \sum_{i=1}^n (x_i - \bar x)^2 = n\bar x (1 - \bar x) $ (as we can see in an answer to this question). So the formula of the sample standard deviation would be $ S = \sqrt {\frac {n} {n-1} \bar x (1 - \bar x)}$.
But in all resources I've read about confidence intervals for proportions, when the population proportion p is unknown, the standard deviation of the population is estimated by $ \sqrt{\bar p (1-\bar p)} $ . This approximation, however, does not use the Bessel correction. If it were, σ would be approximated by $ \sqrt{\frac {n} {n-1} \bar p (1-\bar p)} $.
I understand that $ \bar p (1-\bar p)$ is a consistent estimator for $p(1-p)$, but wouldn't $\frac {n} {n-1} \bar p (1-\bar p)$ be consistend and unbiased, and thus a better estimator?