$\mathbb{Q}(\sqrt{1-\sqrt{2}})$ is Galois over $\mathbb{Q}$

I am trying to show that $E = \mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ is galois over $\mathbb{Q}$. The extension has the minimal polynomial

$$\left(x-\sqrt{1-\sqrt{2}}\right)\left(x+\sqrt{1-\sqrt{2}}\right)\left(x-\sqrt{1+\sqrt{2}}\right)\left(x+\sqrt{1+\sqrt{2}}\right)$$

but I can't manage to show using elementary arithmetic that $\sqrt{1+\sqrt{2}}$ is in $E$ (i.e. multiplying, adding etc.) so that its the splitting field.

The trick would be to use that $\sqrt{1+\sqrt{2}} = \frac{i}{\sqrt{1-\sqrt{2}}}$ but we only know that $i\sqrt{\sqrt{2}-1}$ is in the field, we dont know if we have $i$.

Can someone give a hint on how to proceed, maybe there is a theorem I could use that I'm not thinking of.

The given field $E = \mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ is not a galois extension. If it were it would contain $\sqrt{1 + \sqrt{2}}$ as stated in the question and then $\mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ would contain $\mathbb{Q}\left(\sqrt{1+\sqrt{2}}\right)$ which is an extension of degree 4 as is $\mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ (the minimal polynomial is $X^4-2X^2-1$ for both numbers), thus this inclusion would be an equality which is not the case because $\mathbb{Q}\left(\sqrt{1+\sqrt{2}}\right) \subset \mathbb{R}$ and $\sqrt{1-\sqrt{2}} \in \mathbb{C}$ but $\sqrt{1-\sqrt{2}} \notin \mathbb{R} $.