Let $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3},\sqrt[5]{5})$.
How to show that this is a simple extension without the primitive element theorem?
My idea was to show that $\mathbb{Q}(\sqrt2, \sqrt[3]{3}, \sqrt[5]{5})=\mathbb{Q}(\sqrt{2} \cdot \sqrt[3]{3} \cdot \sqrt[5]{5})$.
The field $L:= \mathbb{Q}(\sqrt2, \sqrt[3]{3}, \sqrt[5]{5})$ contains $\sqrt 2, \sqrt[3]{3}$ and $\sqrt[5]{5}$.
So $\mathbb{Q}(\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5}) \subset L$.
To show that $L \subset \mathbb{Q}(\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5})=:M$, I tried to indicate that $\sqrt2, \sqrt[3]{3}, \sqrt[5]{5} \in M$.
$\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5} \in M \Rightarrow (\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5})^2 \in M$
But now I don't know how to continue.
How to show that this is a simple extension?