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I tried an inductive proof but it didn't work out and I'm stuck for ideas.

user107921
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    When $n=0$, the left hand side is $3$, and the right is $2$. When $n=1$, the left hand side is $9$, and the right is $4$. Etc. What you are trying to prove is false. – Andrés E. Caicedo Nov 11 '13 at 02:07

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This is false, in general: For we have

$$2^{2^{n + 1} - 1} = 2^{2^{n} \cdot 2 - 1} = \frac{1}{2}\left(2^{2^n}\right)^2 > 2^{2^n}$$

for every $n \ge 1$.