Sampling distribution of sample variance 
Let $(X_1, X_2, ..., X_n)$ be a random sample from Bernoulli($p$). Find the sampling distribution of $S_n^2=\frac{\sum_{i=1}^n(X_i -\bar X)^2}{n-1}$.

I've just started learning about sampling distributions of statistics. I have found out that $\sum_{i=1}^nX_i, \sum_{i=1}^nX_i^2$ both follow Binomial($n,p$). Expanding $S_n^2$, I got $\frac{\sum_{i=1}^nX_i^2 + n\bar X^2 - 2\bar X\sum_{i=1}^nX_i}{n-1}$ and this is where I'm stuck.
Please help.
 A: Everything falls into place when you look at the data.
Notice that the $X_i$ are all just zeros or ones.  Suppose $K$ of them are equal to $1,$ so that the remaining $n-K$ are equal to $0.$  This enables you to reduce the sum of $n$ terms into a sum of just two terms,
$$\bar X = \frac{1}{n}\sum_{i=1}^n = \frac{1}{n}\left[K(1) + (n-K)(0)\right] = \frac{K}{n}.$$
Therefore there are only two values of the residuals: for $K$ of the data, $X_i - \bar X = 1 - k/n = (n-K)/n$ and for the remaining data, $X_i - \bar X = 0 - K/n = -K/n.$  Consequently
$$\sum_{i=1}^n \left(X_i - \bar X\right)^2 = K\left(\frac{n-K}{n}\right)^2 + (n-K)\left(\frac{-K}{n}\right)^2 = \frac{K(n-K)}{n}.\tag{*}$$
Next, observe that the random variable $K$ follows a Binomial$(n,p)$ distribution: that is, for each $k\in\{0,1,2,\ldots, n\},$ $$\Pr(K=k) = \binom{n}{k}p^k(1-p)^{n-k}.\tag{**}$$
Formula $(*)$ lists the possible values of $(n-1)S_n^2$ while formula $(**)$ gives their probabilities: thus you have full information about the distribution of $S_n^2$ (which is $1/(n-1)$ times $(*)$).  You could simplify it a little bit by combining the probabilities for any groups of $K$ for which $K(n-K)/n$ have a common value.
I recommend working through this in detail -- looking at all possible (unordered) samples -- for a couple of small values of $n,$ such as $n=2$ and $n=3.$  (The simplification varies a little depending on the parity of $n.$)
