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, $u\ne0$.