Why is the image of a $\pmod p$ Galois representation finite? Let $\overline{\mathbb{F}_q}$ be the algebraic closure of the finite field on $q=p^r$ elements, and $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ the absolute Galois group with the profinite topology. 
Why is it true that every continuous homomorphism,
$$\rho \colon \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{GL}_n(\overline{\mathbb{F}_q})$$ has finite image? 
I understand that in the case of complex representations, $\rho \colon \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{GL}_n(\mathbb{C})$, one can use the fact that there exists open neighbourhoods of $I_n\in \text{GL}_n(\mathbb{C})$ which contain no non-trivial subgroups. This means that $\rho$ has an open kernel, which implies that the image is finite. But I don't think the same method will work for the $\pmod p$ case.
Thanks,
 A: The algebraic closure $\overline{\mathbb{F}_q} = \overline{\mathbb{F}_p}$ is naturally the direct limit of all the intermediate extensions:
$$ \overline{\mathbb{F}_p} = \varinjlim_n \mathbb{F}_{p^n} $$
In particular there are compatible maps $\mathbb{F}_{p^n} \to \overline{\mathbb{F}_p}$ which in this case can be regarded just as inclusions. Hence the natural topology to use on $\overline{\mathbb{F}_p}$ is the one induces by all these maps. As the finite fields have the discrete topology, it is clear that $\overline{\mathbb{F}_p}$ has it as well.
Now, $GL_n(\mathbb{F}_p)$ inherits its topology from the product topology on $\overline{\mathbb{F}_p}^{n^2}$, but this is the discrete one. Therefore, once again we are left with the discrete topology on the general linear group. In particular, the set $\{\mathbb{1}_n\}$ is open (and also closed), so its preimage must be open. This is exactly $\ker \rho$, and by definition of the Krull topology on the Galois group it contains a set of the form $\mathrm{Gal}(\overline{\mathbb{Q}}/K)$ with $K$ a number field. This shows that $\rho$ factors through $K$.
A: Assuming $GL_n(\overline{\mathbb{F}}_q)$ is discrete, note that $\ker \rho= \rho^{-1}(1)$ is open in $G_\mathbb{Q}$. By Galois Theory, it follows that $L=\overline{\mathbb{Q}}^{\ker \rho}$ is a finite Galois extension of $\mathbb Q$ and $G_L=\ker \rho$. Hence 
$
\text{im}\, \rho\cong \text{Gal}(L/\mathbb{Q})
$
is finite. 
