Let $C$ be a $[n,k]_q$ code. Then: $d = n - k + 1$ iff the generator matrix $G$ contains $k$ linear independent columns, meaning that choosing $k$ arbitrary columns of $G$ always leads to $k$ linear independent vectors.
I thought a while about this, but I can't come up with anything specific. $d$ is the minimum distance that two vectors of $C$ can possess, respectively the minimal amount of components $v_i$ of a vector $v \in C$ that are different from $0$. I think since there are $k$ columns that are linear independent from each other, then the code words can differ at least at $k$ components from each other.
It migt be wise to use:
$C$ possesses the minimal distance $d$ iff there are always $d-1$ columns of the check matrix $H$ that are linear independent, but $d$ columns that are linear dependent.
Applying this to our given problem, the theorem could be rewritten as:
$C$ possesses the minimal distance $n - k + 1$ iff there are always $n - k$ columns of the check matrix $H$ that are linear independent, but $n - k + 1$ columns that are linear dependent.
Note that $H = ((-G')^T|E_{n-k})$.
First, notice that $d$ cannot exceed $n-k+1$ (singleton bound), hence $d = n-k+1$ is equivalent to $d \ge n-k+1$
If any $k$ subset of columns of $G$ is LI, then $d \ge n-k+1$. Suppose this were false; then $d\le n-k$, i.e., there exists a (non zero) codeword $c_0$ with at least $k$ zeros. But this cannot be. Because $c_0 = uG$ and any $k\times k$ submatrix of $G$ is full rank (that is, not only the columns are LI, but also the rows).
The converse is similar. Suppose $d \ge n-k+1$. Then there cannon exist a $k$ subset of columns of $G$ that is LD. Because then the respective $k\times k$ submatrix would be singular, hence also its rows would be LD, and then there would exist some $c =uG$ that produces at least $k$ zeros in the codeword.