So $V$ is an $n$-dimensional $K$ vector space. This fact you want to prove is true for any field $K$.
Since $f$ is nonzero, its range must be $K$. Therefore the nullspace of $f$ has dimension $n-1$ by the rank-nullity theorem.
Let $e_2,\ldots,e_n$ be a basis of this nullspace.
Then pick any $e_1$ such that $f(e_1)=1$.
It follows that $e_1,e_2,\ldots,e_n$ is a basis of $V$.
Indeed, if you take a linear combination of the latter and apply $f$, you'll see that the coefficient of $e_1$ is equal to $0$. Then you can use the linear independence of $e_2,\ldots,e_n$ to get the nullity of the other coefficients. So $e_1,\ldots, e_n$ are linearly independent. Given the dimension, they must form a basis of $V$.
as you wanted.