Irrational numbers, induction I have $\sqrt[3]{2}^{2^n}$. 
Can I prove that this number is irrational by showing that $3$ does not divide $2^n$?

If $\dfrac{p}{q} = 2^{\frac{2^n}{3}}, (p,q) = 1 \Rightarrow \dfrac{p^3}{q^3} = 2^{2^n}\Rightarrow p^3=2^{2^n}q^3\Rightarrow 2\mid p^3\Rightarrow 2\mid p$. Now if $p = 2^k$, then $p^3 = 2^{3k} \Rightarrow q^3 = 2^{2^n-3k}$. Observe $3\nmid 2^k \Rightarrow 2^n - 3k \geq 1 \Rightarrow 2\mid q^3 \Rightarrow 2\mid q \Rightarrow (p,q) \geq 2 \neq 1$, contradiction. Thus if $p = (2r+1)2^k \Rightarrow p^3 = (2k+1)^3\cdot 2^{3k} = 2^{2^n}q^3$. Now if $2^{3k} > 2^{2n} \Rightarrow q$ is even. Thus $(p,q) \geq 2 \neq 1$, contradiction. And if $2^{3k} < 2^{2^n}$, then $(2k+1)^3$ is even, contradiction, since $2k+1$ is odd, so is its cube power. Thus $\sqrt[3]{2^{2^n}}$ is irrational. 
