Prove that if $\mathbb Q=K_0 \subset K_1... \subset K_n$ is a sequence of fields such that $K_{i+1}/K_i$ are Galois extensions and $[K_{i+1}:K_i]=3 $ for each $i$, then $\mathbb Q(2^{1/3})$ is not contained in $K_n$. I just want to check if my attempt is correct.
Assume it is true, let $a= 2^{1/3}$ and $f(x)=x^3-2 \in K_{n-1}$. Then as $K_n/K_{n-1}$ is Galois and f has a root in $K_n$, therefore f splits in $K_n$ ($f(a)=0$). The roots of f are $a, \omega a,\omega^2 a$ where $\omega$ is a third root of unity. Therefore $\omega=e^{i2\pi /3}\in K_n$ and thus $\mathbb Q \subset \mathbb Q[\omega] \subset K_n$. However, $[\mathbb Q[\omega]:\mathbb Q]=2$ so $2$ divides $[K_n: \mathbb Q]=3^n$ (tower law) which is a contradiction.
I didn't use the fact that $K_{i+1}/K_i$ is Galois apart from the case $i=n-1$ so I am not sure if this is right.
Minor comment: You claim that $f$ splits in $K_n$ because it has a root in $K_n$. This uses the fact that $f$ is irreducible over $\Bbb{Q}$, which you do not mention.
Major problem: How do you conclude that $i\in K_n$? This seems to come out of the blue.
As for your edited question; yes, your argument is now fine.