$\displaystyle {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2$
There is not much "simplification" to be done, but you may want to write one of the following:
$\displaystyle {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2
= {1\over n}\sum_{i=1}^n \left(X_i - \frac{1}{n}\sum_{j=1}^n X_j\right)^2
$
(which you get by expanding to square in the right-hand side into $\displaystyle X_i^2 - 2X_i{1\over n}\sum_{j=1}^n X_j + \left({1\over n}\sum_{j=1}^n X_j\right)^2$). It is the other expression for the empirical variance.
or:
$\displaystyle {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2
= {1\over n}\sum_{i=1}^n X_i^2 - {1\over n^2}\sum_{1\leq i, j\leq n}X_iX_j$$\displaystyle ={1\over n}\sum_{i=1}^n X_i^2 - {1\over n^2}\sum_{i=1}^n X_i^2 - {1\over n^2}\sum_{1\leq i\neq j\leq n} X_iX_j = \left(\frac{1}{n}-\frac{1}{n^2}\right)\sum_{i=1}^n X_i^2 - \frac{2}{n^2} \sum_{1\leq i<j \leq n} X_i X_j$
(which is the expansion of the square of the sum. In the last expression, every product $\displaystyle X_iX_j$ appears exactly once)
I hope it answers you question.
Laurent.