Applying the additivity of inner product in each argument and orthogonality of the collection $x_1, \ldots, x_N$, we have \begin{equation} \begin{split} \left\Vert \sum_{n = 1}^N x_n \right\Vert^2 & = \left( \sqrt{\left\langle \sum_{n = 1}^N x_n, \sum_{n = 1}^N x_n \right\rangle} \right)^2 \\ & = \left\langle \sum_{n = 1}^N x_n, \sum_{n = 1}^N x_n \right\rangle \\ & = \sum_{n = 1}^N \sum_{m = 1}^N \langle x_n, x_m \rangle = \sum_{n = 1}^N \sum_{m = 1}^N \langle x_n, x_m \rangle I_{\{ n \}}(m) = \sum_{n = 1}^N \langle x_n, x_n \rangle = \sum_{n = 1}^N \Vert x_n \Vert^2 \end{split} \end{equation} where the $I_{\{ n \}}$ are each an D41: Indicator function. $\square$