What is the degree of the field extension $\mathbb Q(\sqrt 2, \sqrt[4] 2,\sqrt[8] 2)$ over $\mathbb Q$?

Question

What is the degree of the field extension $\mathbb Q(\sqrt 2, \sqrt[4] 2,\sqrt[8] 2)$ over $\mathbb Q$?

The options given are $4$, $8$, $14$, $32$. Which one is correct?

I think the answer is $8$. What I have done is to use the following $$\begin{align} [ \mathbb Q(\sqrt 2, \sqrt[4] 2,\sqrt[8] 2) : \mathbb Q] &= [\mathbb Q(\sqrt 2, \sqrt[4] 2,\sqrt[8] 2) : \mathbb Q(\sqrt 2, \sqrt[4] 2)] \cdot [\mathbb Q(\sqrt 2, \sqrt[4] 2) : \mathbb Q(\sqrt 2) ]\cdot[\mathbb Q(\sqrt 2) : \mathbb Q ].\end{align}$$ and conclude that each has a degree $2$ on the right hand side, thus total degree is $8$.

Am I right?

Answer 1

Hint

$$\left(\sqrt[8]2\right)^2=\sqrt[4]2$$ and therefore $$\mathbb Q(\sqrt 2, \sqrt[4] 2,\sqrt[8] 2)=\mathbb Q(\sqrt 2,\sqrt[8] 2)$$