Proof by induction for $f^{n+1}(x)=\frac{x}{\sqrt{1+(n+1)x^2}}$

Question

Look at the following function f: $\mathbb{R} \to \mathbb{R}: x \mapsto \frac{x}{\sqrt{1+x^2}}.$

Show with the complete induction that the recursive ( given by $f^1:=f$ and $f^{n+1}:=f\circ f^n$) composition $f^n$ from $f$ has the following explicit illustration:

$f^n(x)=\frac{x}{\sqrt{1+nx^2}}$

My solution:

Base case: For $n=1$

$f^1(x)=f(x)=\frac{x}{\sqrt{1+x^2}}=\frac{x}{\sqrt{1+1\cdot x^2}}$

Inductive step:

$f^{n+1}(x)=(f\circ f^n)(x)=f(f^n(x))=\frac{f^n(x)}{\sqrt{1+(f^n(x))^2}}$

So now I would have to plug in $f^n(x)$ for $x$ I think.

$\frac{x}{\sqrt{1+nx^2}}$ $\frac{\frac{x}{\sqrt{1+nx^2}}}{\sqrt{{1+(\frac{x}{\sqrt{1+nx^2}}})^2}}$ But how do I get from there to $\frac{x}{\sqrt{1+(n+1)x^2}}$ = $\frac{x}{\sqrt{1+nx^2+x^2}}$

I have encountered this problem in an old math exam from 2016 and it is of interest to me because I am currently practicing a lot of exercises related to convergences and mathematical induction.

Any hints guiding me to the right direction I much appreciate.

Answer 1

\begin{align} (f\circ f^n)(x)&= \dfrac{ \dfrac{x}{\sqrt{1+nx^2}} }{\sqrt{1+\dfrac{x^2}{1+nx^2}}}\\ & = \dfrac{ \dfrac{x}{\sqrt{1+nx^2}} }{\sqrt{\dfrac{1+nx^2+x^2}{1+nx^2}}}\\ & =\dfrac{x}{ \sqrt{1+nx^2+x^2}}\\ &=\dfrac{x}{ \sqrt{1+(n+1)x^2}}\\ &= f^{n+1}(x) \end{align}