Proof that there is a unique solution to $Ax=b$ implies rank = $n$ of a $m X n$ matrix $A$

Proof that there is a unique solution to $Ax=b$ implies rank = $n$ of a $m
X n$ matrix $A$

I am confused in some part of the proof. Let $A =[\bar{a_1}
\bar{a_2}\dots\bar{a_n}]$. Suppose, rank($A$) < $n$ then there is a $k$
such that $$a_k = \sum_{i=1}^{k-1}\alpha_{i} \bar{a_i}$$
Now suppose $(x_1, x_2,\dots,x_n)$ is a solution. Then we can generate
another solution vector where the first $k-1$ terms are $(x_i + \alpha_i
x_{k})$ where $i=1,2,\dots,k-1$, the $k$-th term is $0$ and the rest of
the terms are as it is. Now, if $x_k$ is non-zero I am getting two
different solution vectors, thus getting contradiction, but if $x_k$ is
zero in the original solution we get the same vector by our construction
process. So, how to get a contradiction ?