Original question: Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x)=\dfrac{a^x-a^{-x}}{2}$, where $a>0$ and $a\ne 1$, and $\alpha$ be a real number such that $f(\alpha)=1$. Find $f(2\alpha)$.$^1$
A few years ago, I was a high school student and solved it. Now I am reading the book again, because I has begun teaching my cousin since last week. When I revisited the question, suddenly I wanted to find $f(2\alpha)$, $f(3\alpha)$, $f(4\alpha),\;\dots$ \begin{align} f(2\alpha)&=\frac{a^{2\alpha}-a^{-2\alpha}}{2}\\ &=\frac{(a^{\alpha}-a^{-\alpha})(a^{\alpha}+a^{-\alpha})}{2}\\ &=f(\alpha)\sqrt{a^{2\alpha}+2+a^{-2\alpha}}\\ &=f(\alpha)\sqrt{(a^{\alpha}-a^{-\alpha})^2+4}\\ &=f(\alpha)\sqrt{4(f(\alpha))^2+4}\\ f(3\alpha)&=f(\alpha)(4(f(\alpha))^2+3)\;(\text{calculations skipped})\\ f(4\alpha)&=f(\alpha)(4(f(\alpha))^2+2)\sqrt{4(f(\alpha))^2+4} \end{align}
My question: Can we express $f(n\alpha)$ in terms of $f(\alpha)$? (Here $f(\alpha)$ isn't necessarily $1$.)
Attempt: It is known that $x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+y^{n-1})$ for $n\in \mathbb{N}$, so $$f(n\alpha)=\frac{a^{n\alpha}-a^{-n\alpha}}{2}=\frac{(a^{\alpha}-a^{-\alpha})(a^{(n-1)\alpha}+a^{(n-3)\alpha}+\cdots+a^{(3-n)\alpha}+a^{(1-n)\alpha})}{2}.$$ However, $(a^{(n-1)\alpha}+a^{(n-3)\alpha}+\cdots+a^{(3-n)\alpha}+a^{(1-n)\alpha})$ term is annoying me. I think it will be $2(f((n-1)\alpha)+f((n-3)\alpha)+\cdots+?)$, but I have no idea how to do next.
Partial solution is also appreciated.
$^1$ It was translated from Korean to English by me. Reference: Sunwook Hwang and 12 other authors (2010).『수학Ⅰ 익힘책』. Seoul: (주)좋은책신사고. page 62.