I am trying to prove this. For me the definition of basis was presented without the condition that each vector in $\mathbb{R}^n$ is a linear combination of the basis vectors. That result was derived later using the result, that a basis of $\mathbb{R}^n$ includes exactly $n$ vectors. This however isn't clear to me. Here's my attempt to prove it:
I assume, that there are m vectors in $\mathbb{R}^n$ in this linearly independent set:
$$ \sum_{k=1}^m\xi_kx_k=0, \ \ x_k=\left(\begin{matrix} x_{k1}\\ x_{k2}\\ \vdots\\ x_{kn} \end{matrix}\right) $$ This can be written as a matrix equation: $$ \Leftrightarrow\ \underbrace{\left(\begin{matrix} x_{11}&x_{21}&\dots&x_{m1}\\ x_{12}&x_{22}&\cdots&x_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ x_{1n}&x_{2n}&\dots&x_{mn} \end{matrix}\right)}_{=A} \underbrace{\left(\begin{matrix} \xi_1\\ \xi_2\\ \vdots\\ \xi_m \end{matrix}\right)}_{=x}=0 $$ Now the question is, when is the solution vector $x$ unique? When this is answered, it should follow that $m = n$.
It's false. One vector is linearly independent in any $\Bbb R^n$, for $n\gt1$.
Pretty clear that you meant in a basis.
For the solution to exist and be unique, in your setup, you need the matrix to be invertible. In which case it's square, and $m=n$.