How many different bases in $\mathbb{Z}/p\mathbb{Z}$ Let $K = \mathbb{Z}_p$, for some prime $p$, and $\text{dim}\:V = n$. $V$ is a vector space over $K$.
I need to find out how many different bases are in $V$.
Now I know the answer is the product of all 
$$\frac{1}{n!}\prod\limits_{i=0}^{n-1} (p^n-p^i)$$
The solution states that there are $p^n-1$ choices for the first vector. Why is this? I can see why there will be $p^n$ choices but why is it $-1$ ? 
My thoughts are that we are taking away the zero vector, but i'm not sure if this is the reason why.

$-1$ because a basis can't contain the zero vector. ;) More precisely, a family of vectors which contains the zero vector $\mathbf{0}$ can't be free, because then you have a linear combination with non-zero coefficients which yields the zero vector (e.g., $1\cdot \mathbf{0} = \mathbf{0}$ ).
