A simple proof for non-rationality of number $\sqrt[n]{1+q^n}$

Question

I'm going to prove the following fact (preferably with an elementary simple method):

Let $n>2$ be a natural number. Then for all $q \in \mathbb{Q^+}\setminus\lbrace 0 \rbrace$ we have $\sqrt[n]{1+q^n} \not\in \mathbb{Q^+}\setminus\lbrace 0 \rbrace$.

Assuming the contrary, we then have $(1+\frac{a^n}{b^n})=\frac{c^n}{d^n}$, where $q=\frac{a}{b}$ and $\sqrt[n]{1+q^n}=\frac{c}{d}$. Now multiplying both sides of the above equation by $b^nd^n$ we get a contradiction according to Fermat Last Theorem. But the problem doesn't seem to be that difficult that using the Fermat theorem is its unique solution. So I'm looking for a simpler alternative proof which does not use the Fermat theorem. Would you please guide me for a simple argument?
Thanks in advance.

Answer 1

If you have $a^n+b^n=c^n$ for non-zero integers $a,b,c$, then $$\sqrt[n]{1+\frac{b^n}{a^n}}=\frac{c}{a}$$so a proof of your statement would prove FMT. This means there is no shortcut.