I've seen this exercise in several statistics text, but how they get to the final formula is something that I don't quite get. How do two squared terms suddenly become a binomial term? I've been trying to figure out how to get to the final formula but I don't get anywhere near. Hope you can help me.
On
$$\begin{array}{rl} n \|\mathrm x - \bar{x} \, 1_n\|_2^2 &= n (\mathrm x^T \mathrm x - 2 \bar{x} 1_n^T \mathrm x + n \bar{x}^2)\\\\ &= n (\mathrm x^T \mathrm x - 2 n \bar{x}^2 + n \bar{x}^2)\\\\ &= n (\mathrm x^T \mathrm x - n \bar{x}^2)\\\\ &= n \|\mathrm x\|_2^2 - (n \bar{x})^2\\\\ &= n \|\mathrm x\|_2^2 - (1_n^T \mathrm x)^2\end{array}$$
One may recall that $$ \bar{x}=\frac1N\cdot \sum_{n=1}^Nx_n $$ giving $$ \begin{align} N^2\cdot \frac1N\sum_{n=1}^N(x_n-\bar{x})^2&=N\sum_{n=1}^N(x_n^2-2x_n\cdot\bar{x}+\bar{x}^2) \\\\&=N\sum_{n=1}^Nx_n^2-2\:N\cdot\bar{x}\sum_{n=1}^Nx_n+N\cdot N\cdot \bar{x}^2 \\\\&=N\sum_{n=1}^Nx_n^2-2\:\left(\sum_{n=1}^Nx_n\right)^2+\left(\sum_{n=1}^Nx_n\right)^2 \\\\&=N\sum_{n=1}^Nx_n^2-\left(\sum_{n=1}^Nx_n\right)^2 \\\\&=\sum_{n=1}^Nx_n^2\cdot\sum_{n=1}^N1-\sum_{n=1}^Nx_n \cdot \sum_{n=1}^Nx_n \\\\&=\det M \end{align} $$ as announced.