Cardinality of $GL_n(K)$ when $K$ is finite I don't know how to do the last task of an exercise.  
Let $K$ be a field, $G=GL_n(K)$ and $X=K^n\backslash\{0\}$.
First task: Show that $G \times X \to X$, $(A,x)\mapsto Ax$ defines an action of $G$ on $X$. Done.
Second task: find the stabilizer of $\displaystyle x=\left(\begin{eqnarray} 1\\ 0 \\ \vdots \\0 \end{eqnarray}\right)$. Done.
Third task: Show that there is only one orbit. Done.
Last task: Let $|K|<\infty$. Use the last two tasks and induction to find $|G|$.  
I know that $|G|=[G:St_G(x)]\cdot|St_G(x)|$ and i know that $[G:St_G(x)]=|\bar{x}|=m^n-1$ but i can't figure out what $|St_G(x)|$ is.
What should i do next?
I appreciate every hint :)

I think we can close to the last part like this. You know that if $V=V_n(q)$ be a vector space on the field $F=GF(q)$ of dimension $n$ then we have vectors $e_i, ~1\leq i\leq n$ such that $$V=\langle e_1,e_2,...,e_n\rangle$$ So $|V|=q^n$. In fact, $V\cong F\times F\times...\times F$ $n$ times. And you know that $$GL_n(q)=\{T\mid T:V\to V\ ~\text{where} ~~T~~\text{are all invertible linear transformations}\}$$ Now take $T\in GL_n(q)$. What is the order of $T(e_1)$? Obviously it is $q^n-1$. What will be $T(e_2)$? It is clearly $q^n-q$. By this way we will have $T(e_n)=q^n-q^{n-1}$. This gives you $$|GL_n(q)|=(q^n-1)(q^n-q)...(q^n-q^{n-1})$$
