Showing $ \mathbb{Q}[\sqrt[4]{2}]$ is a field Here i want to prove
\begin{align}
  \mathbb{Q}[\sqrt[4]{2}] = \{ a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8} \mid a,b,c,d\in\mathbb{Q}\} 
\end{align}
this is a field. To do that i want to prove
\begin{align}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} \in \mathbb{Q}[\sqrt[4]{2}]
\end{align}
Is there any ideas for showing this?

[edit] Many of you prove $\mathbb{Q}[\sqrt[4]{2}]$ is a field in indirect way. Is there any ways for showing this in direct way? For $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt[3]{2}]$, using some formulas i can check that indeed every nonzero elementn is unit. But this case does not seem that easy...

Idea: use that $\sqrt[4]{2} = \sqrt{ \sqrt{2}}$, so first rationalize to a denominator in $\mathbb{Q}(\sqrt{2})$, and then finish off.
$$\frac{1}{a+b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}}= \frac{1}{a + c \sqrt{2} + (b + d \sqrt{2}) \sqrt[4]{2}}= \\
=\frac{a + c \sqrt{2} - (b + d \sqrt{2}) \sqrt[4]{2}}{(a + c \sqrt{2})^2 -\sqrt{2}(b + d \sqrt{2})^2}$$ and rationalize by amplifying the fraction by $(a - c \sqrt{2})^2 +\sqrt{2}(b - d \sqrt{2})^2$
