Let $T = \{\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7}\}$ and $\mathcal{S}$ the set of all functions that maps $T$ into $\mathbb{Q}.$ What is the cardinality of $\mathcal{S}$?
So far I've been messing with the idea that we can take the subset $T_0\subset T$ of functions that square an element of $T$ and adds any other string of rational numbers to it. For example $f(\sqrt{2})=(\sqrt{2})^2+a$ where $a \in \mathbb{Q}.$ Since $|T_0|$ is countable and infinite $|T_0|=\aleph_0.$
I'm not really sure if this is going to get me anywhere.
Could you answer the question of the cardinality of the set of all functions from $\{ 1 \}$ to $\mathbb{Q}$? I hope so, it's obviously the same as $\mathbb{Q}$ since any such function can be specified by a single value from $\mathbb{Q}$.
Now, how about the cardinality of the set of all functions from $\{ 1, 2 \}$ to $\mathbb{Q}$? Well, any such function can be specified by two values from $\mathbb{Q}$ so its cardinality is the same as $\mathbb{Q} \times \mathbb{Q} = \mathbb{Q}^2$. Do you know the cardinality of this set?
Let's jump a step to the set of functions from $\{1, 2, 3, 4 \}$ to $\mathbb{Q}$. As Seewoo says, this has cardinality $\mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} = \mathbb{Q}^4$. If you figured that $\mathbb{Q}^2$ is countable then you should see that this is as well.
Now your set: $\{\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7}\}$. Well there is a very simple bijection between it and my set $\{1, 2, 3, 4 \}$ which can be easily used to associate the maps from my set to $\mathbb{Q}$ with those from your set to $\mathbb{Q}$.