Given $n-1$ linearly independent vectors, $\{v_j\}_{j=1}^{n-1}$ in $\mathbb{R}^n$, we can find a non-zero vector, $u$, perpendicular to all of them.
If we set $ \begin{align} u_1&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&1\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&0 \end{bmatrix}\\ u_2&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&0\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&0 \end{bmatrix}\\ &\vdots\\ u_n&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&0\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&1 \end{bmatrix}\\ \end{align} $ then $ u\cdot w=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&w_1\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&w_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&w_n \end{bmatrix} $ If we replace $w$ by any of the $v_j$, the determinant will be $0$ because of duplicate columns; thus, $u\cdot v_j=0$.
$\{v_j\}_{j=1}^{n-1}$ cannot span $\mathbb{R}^n$, so there must be some $v_n$ that is not in the span of $\{v_j\}_{j=1}^{n-1}$. This means that $\{v_j\}_{j=1}^n$ are independent, and so $ \begin{align} u\cdot v_n&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&v_{n,1}\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&v_{n,2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&v_{n,n} \end{bmatrix}\\ &\ne0 \end{align} $ In particular, $u\ne0$.