Norm of elements in number field are rational

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Let $K$ be a finite extension of $\mathbb{Q}$, and let $Hom(K,\mathbb{C})$ be the set of $\mathbb{Q}-$homomorphisms from $K$ to $\mathbb{C}$.

Define the norm of $a\in K$ to be $$N(a)=\prod_{\sigma \in Hom(K,\mathbb{C})}\sigma(a)$$

And we want to prove that $N(a)$ is a rational number.

I saw a proof somewhere but I couldn't understand it well:

A possible proof

Suppose $\tau\in Gal(\mathbb{C}/\mathbb{Q})$

Since by replacing every $\sigma(a)$ by $\tau\sigma(a)$ we get a permutation of the product in the definition of $N(a)$, so we can easily have $$\tau(N(a))=\prod_{\sigma}\tau\sigma(a)=N(a)$$

By this we have $N(a)\in \mathbb{Q}$

My confusion about this proof

I have no idea about the fixed field of the galois group $Gal(\mathbb{C}/\mathbb{Q})$, so I don't know why we can deduce $N(a)\in \mathbb{Q}$ from the above equation.

My question is: why is the above "proof" valid or not?

Thank you very much!