Finding the ring of integers of $\Bbb Q(\sqrt[4]{2})$ 
I know$^{(1)}$ that the ring of integers of $K=\Bbb Q(\sqrt[4]{2})$ is $\Bbb Z[\sqrt[4]{2}]$ and I would like to prove it.

A related question is this one, but it doesn't answer mine.
I computed quickly the discriminant $\text{disc}(1,\sqrt[4]{2},\sqrt[4]{4},\sqrt[4]{8})=-2^{11}$.
According to this answer, this means that
$\mathcal{O}_K \subset \frac{1}{m}\Bbb Z\left[\sqrt[4]{2}\right]$ where $m$ is an integer whose square divides $2^{11}$, so $m=1,2,2^2,\dots,2^5$ are possible.
But how could I rule out the values $m>1$?
I am aware that it can be a tricky problem. Any reference providing a description of $\mathcal{O}_{\Bbb Q(\sqrt[4]{2})}$ would be satisfactory. I will be grateful for any help!

$^{(1)}$ I tested with SAGE the following code
K.<a> = NumberField([x^4-2]);
K.integral_basis()

and I got the expected answer, namely $[1,a,a^2,a^3]$.

There is a quick argument available in this case. Write $\alpha=\sqrt[4]{2}$ and $K=\mathbb{Q}(\alpha)$.
A prime number $p$ divides the index $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ if and only if there is a non-invertible prime $\mathfrak{p}$ of $\mathbb{Z}[\alpha]$ lying over $p$. In this case we have
$$
-2^{11}=\Delta(\mathbb{Z}[\alpha])=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\cdot\Delta(\mathcal{O}_K),
$$
so to prove that $[\mathcal{O}_K:\mathbb{Z}[\alpha]]=1$, it suffices to show that every prime $\mathfrak{p}$ of $\mathbb{Z}[\alpha]$ lying over $2$ is invertible.
If $\mathfrak{p}$ is such a prime, we have $\alpha\in\mathfrak{p}$ because $\alpha^4=2\in\mathfrak{p}$. As $(\alpha)$ is already a prime ideal (note that $\mathbb{Z}[\alpha]/(\alpha)\cong\mathbb{F}_2$), we have $(\alpha)=\mathfrak{p}$. But then the identity $(2)=\mathfrak{p}^4$ proves that $\mathfrak{p}$ is invertible; its inverse is the fractional ideal $2^{-1}\mathfrak{p}^3$.
