Find the number of surjective linear transformations 
Let $V$ and $W$ be vector spaces over a finite field $F$ of order $q$ and $m=\dim(V)\geq \dim(W)=n$. Find the number of surjective linear transformations from $V$ to $W$.

I know that if $B=\{v_1,\dotsc,v_m\}$ is a basis for $V$ and if $T:V\to W$ is linear, then $T$ is surjective if and only if $T(B)$ generates $W$. Clearly $|T(B)|\leq m$ and and since $T(B)$ generates $W$ then $|T(B)|\geq n$. Thus it enough to find the number of generators $G$ for $W$ such that $n\leq|G|\leq m$. But how to find such generators? I need help.
 A: A surjective linear map $V\to W$ is precisely determined by (1) its kernel$~K$, which is a $n-m$-dimensional linear subspace of the $n$-dimensional space $V$, and (2) an isomorphism $V/K\to W$.
The basic ingredient to do the counting is a formula for the number of linearly independent families of $k$ vectors in $F^n$. By simple counting of choices this number is
$$
  (q^n-1)(q^n-q)\ldots(q^n-q^{k-1})=q^\binom k2(q-1)^k[n]_q[n-1]_q\ldots[n-k+1]_q
$$
where $[m]_q=1+q+\cdots+q^{m-1}=\frac{q^m-1}{q-1}$ is a $q$-integer. In particular the number of different bases for a vector space of dimension$~m$ is $q^\binom m2(q-1)^m[m]_q[m-1]_q\ldots[1]_q$, which product of $q$-integers is usually written $[m]_q!$. The number of choices for $K$ now is obtained by dividing the number of independent $k$-families but the number of different bases for the same $m-n$ dimensional subspace:
$$
  \frac{q^\binom{n-m}2(q-1)^{n-m}[n]_q[n-1]_q\ldots[m+1]_q}
       {q^\binom{n-m}2(q-1)^{n-m}[n-m]_q!}
 =\frac{[n]_q[n-1]_q\ldots[m+1]_q}{[n-m]_q!}
 =\frac{[n]_q!}{[n-m]_q![m]_q!}
$$
which is the Gaussian binomial coefficient usually written $\binom n{n-m}_q=\binom nm_q$. This gives the answer for (1). The answer for (2) is the number $q^\binom m2(q-1)^m[m]_q!$ of bases of$~W$. Multiply to get the final answer
$$
  \binom nm_qq^\binom m2(q-1)^m[m]_q!
 =q^\binom m2(q-1)^m\frac{[n]_q!}{[n-m]_q!}.
$$
