By simple extension I mean that the exists an element $\alpha\in\mathbb{C}$ such that $\mathbb{C}=\mathbb{Q}(\alpha)$ and thus $\mathbb{C}/\mathbb{Q}=\mathbb{Q}(\alpha)/\mathbb{Q}$. By considering $\alpha$ to be $i$ do we have what we are looking for?
To finish complete the exercise can anyone please confirm that this field extension is in fact algebraic but not finite?
Thank you
It is not algebraic, since there are non-algebraic ie. transcendental numbers in $\Bbb R \subseteq \Bbb C$. The numbers $\pi$ and $e$ come to mind. As far as I know this is not that easy to show. In the case of $\pi$ it is a theorem of Lindemann. @AnneBauval's cardinality argument is quite nice to show that there exist some transcendental elements in $\Bbb R \mid \Bbb Q$. Anyway, using this it is clear that $\Bbb C$ cannot be a simple algebraic extension of $\Bbb Q$, let alone $\Bbb Q(i)$, since the latter is a simple algebraic extension of $\Bbb Q$.
Now let me hijack the question: It could still be the case that $\Bbb C$ is simple in the sense that $\Bbb C \cong \Bbb Q(x)$ for some transcendental element $x$, ie. that $\Bbb C$ is a field of rational functions in one variable over $\Bbb Q$. But again we can use @AnneBauval's trick to deduce that $\Bbb Q(x) \cong \operatorname{Frac}(\Bbb Q[x])$ is a countable field, so out of cardinality reasons $\Bbb C$ cannot even be a simple transcendental extension of $\Bbb Q$.