\begin{align*}
\sigma^2 & = \frac{\sum_{k=1}^n ( x_k-\mu )^2}{n}\\
& = \frac{\sum_{k=1}^n ( x_k^2 - 2x_k \mu + \mu^2 )}{n}\\
& = \frac{\sum_{k=1}^n x_k^2 - 2\mu \sum_{k=1}^n x_k + n \mu^2 )}{n}\\
& = \left ( \frac{\sum_{k=1}^n x_k^2 }{n} \right ) - \mu^2
\end{align*}
The second part is proved similarly. Using the first part of the proof above,
\begin{align*}
\sigma^2 & = \frac{\sum_{k=1}^n ( x_k-\mu )^2}{n}\\
& = \left ( \frac{\sum_{k=1}^n x_k^2 }{n} \right ) - \mu^2\\
& = \left ( \frac{\sum_{k=1}^n x_k (x_k - 1) + x_k }{n} \right ) - \mu^2\\
& = \left ( \frac{\sum_{k=1}^n x_k (x_k - 1)}{n} \right ) + \mu - \mu^2
\end{align*}