Here is the problem: 

Let $K$ be a field with $q$ elements and $V$ be a vector space over $K$. If you'd collect vectors from $V$ (and never took the same one twice) what is the minimum amount of elements which you could be sure to be a spanning set(a generator set) for $V$? 

My reasoning is as follows: 

There must be $q^n$ vectors in $V$, since that is the amount of different combinations of $q$ elements in n value slots (an n-dimensional vector) So of course $q^n$ must be a spanning set. If $(q^n) - 1$ was not a spanning set that would mean that $q^n$ was a minimal spanning set and therefore a base of $V$. But that is impossible since a base of any $n-$dimensional vectorspace contains n elements. Therefore $(q^n)-1$ is a spanning set as well. This applies to $(q^n) -2$ and so on.. Until - we reach $n$ vectors. Therefore every set of vectors containg between $q^n$ and $n$ elements must be a spanning set and therefore the minimal amount of elements you have to collect to have a spanning set for sure is $n$.

But that reasoning contradicts the solution of the problem, which is $(q^n-1) +1. $

Can anyone help me understand my error and the solution?

Looking forward very much to your responses!