How to express $f(n\alpha)$ in terms of $f(\alpha)$

We can rewrite the function as

$$f(x)=\sinh(x\ln a)$$

then $f(nx)=\sinh(nx\ln a)$

let, $x\ln a=t$ then, $f(t)=\sinh(t)$ and $f(nt)=\sinh(nt)$ also, $\cosh(t)=\sqrt{1+f(t)^2}$

Finally use the fact that $$\sinh(nt)=\dfrac{(\cosh(t)+\sinh(t))^n-(\cosh(t)-\sinh(t))^n}{2}$$

From $f(\alpha)=\frac{a^\alpha - a^{-\alpha}}{2}$, we can induce the quadratic equation for $a^{\alpha}$: $$ a^{2\alpha}-2f(\alpha)a^{\alpha}-1=0. $$ By the quadratic formula, we get $$ a^{\alpha}=f(\alpha)\pm \sqrt{(f(\alpha))^2+1} $$ and using $a^{\alpha}>0$ we eliminate one possibility. Thus $$ a^{n\alpha}=(a^{\alpha})^n =\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n $$ and \begin{align} a^{-n\alpha}&=\frac{1}{\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n}\\ &=\frac{\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}{\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}\\ &=\frac{\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}{(f(\alpha)^2+1-f(\alpha)^2)^n}\\ &=\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n. \end{align} Therefore $$ f(n\alpha)=\frac{1}{2}\left(\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n -\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n\right) $$ Checking the formula for $n=2$: \begin{align} f(2\alpha)&=\frac{1}{2}\left(\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^2 -\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^2\right)\\ &=\frac{1}{2}(f(\alpha)^2+1+2f(\alpha)\sqrt{f(\alpha)^2+1} + f(\alpha)^2 -f(\alpha)^2-1+2f(\alpha)\sqrt{f(\alpha)^2+1}-f(\alpha)^2)\\ &=\frac{1}{2}\cdot 4f(\alpha)\sqrt{f(\alpha)^2+1}\\ &=2f(\alpha)\sqrt{f(\alpha)^2+1} \end{align}

Let $a^x = y$, then $y$ can be solved in terms of the RHS. Let $f(\alpha) = A$, from:

$$ y - \frac{1}{y} = 2A $$ with the quadratic root formula, we have $$y = g(A) = A + \sqrt{1 + A^2}$$ After this it is a matter of substitution and simplification: $$ f(n \alpha) = \frac{1}{2} (g(A)^n - g(A)^{-n}) $$