Prove equality concerning differentiable function Let $f(x)$ be a function that is n-times differentiable. Prove the following equality(without L'Hospital):
$$(x^{n-1}\cdot f(\frac{1}{x}) )^{(n)}=\frac{(-1)^{n}}{x^{n+1}}\cdot f^{(n)}(\frac{1}{x})\\$$
So, I wanted to prove it by induction, for $n=0$ the statement is obviously true. So let's assume it's true for some $n \in \mathbb{N_0}$
$\\$
Now we need to check for $n+1$ but I'm not sure how to proceed.
$$(x^{n}\cdot f(\frac{1}{x}) )^{(n+1)}$$
I thought about taking the first derivative of the product, then separately observe the n-th derivative of the two expressions but haven't reached any useful conclusions. Any kind of hint would be appreciated, thanks in advance!

Prove by induction. Let $P_n$ be the claim for the specific value of $n$
Base Case: $$ x^{-1} \cdot f \left( \frac{1}{x} \right) = \frac{1}{x} f \left( \frac{1}{x} \right) $$
$\therefore P_0$
Inductive Step: Assume $P_k$ holds for some whole number $k$, for the sake of induction. By $P_k$ we have:
\begin{align}
\frac{d^k}{dx^k} \left[ x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] &= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right)
\end{align}
Now we note that:
\begin{align}
\frac{d^{k+1}}{dx^{k+1}} \left[ x^{k} \cdot f \left( \frac{1}{x} \right) \right] &= \frac{d^{k}}{dx^{k}} \frac{d}{dx} \left[ x \cdot x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] \\
&= \frac{d^{k}}{dx^{k}}\left[x^{k-1} \cdot f \left( \frac{1}{x} \right) + x \frac{d}{dx} \left[ x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) +\frac{d^{k}}{dx^{k}}\left[x \frac{d}{dx} \left[ x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) +\frac{d^{k}}{dx^{k}}\left[x \left[ (k-1) x^{k-2} \cdot f \left( \frac{1}{x} \right) + x^{k-1}\cdot f' \left( \frac{1}{x} \right) \cdot (-x^{-2}) \right] \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) +\frac{d^{k}}{dx^{k}}\left[(k-1) x^{k-1} \cdot f \left( \frac{1}{x} \right) - x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) + (k-1)\frac{d^{k}}{dx^{k}} \left[x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] - \frac{d^{k}}{dx^{k}} \left[ x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) + (k-1)\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-1}\cdot \frac{1}{x} f' \left( \frac{1}{x} \right) \right] \\ 
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-1}\cdot g\left( \frac{1}{x} \right) \right] \end{align}
where
$$ g(x) = xf'(x) $$
So
\begin{align}
\frac{d^{k+1}}{dx^{k+1}} \left[ x^{k} \cdot f \left( \frac{1}{x} \right) \right] &= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-1}\cdot g\left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{(-1)^k}{x^{k+1}} \cdot g^{(k)} \left( \frac{1}{x} \right)
\end{align}
By the General Leibniz Rule:
\begin{align*}
    \frac{d^k}{dx^k}[g(x)] &= \sum_{m = 0}^k \binom{k}{m}\frac{d^m}{dx^m}[x] \frac{d^{k-m}}{dx^{k-m}}[f'(x)] \\
    &= x f^{(k+1)}(x) + k f^{(k)}(x)
\end{align*}
So
\begin{align*}
\frac{d^{k+1}}{dx^{k+1}} \left[ x^{k} \cdot f \left( \frac{1}{x} \right) \right] &= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{(-1)^k}{x^{k+1}} \cdot g^{(k)} \left( \frac{1}{x} \right) \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{(-1)^k}{x^{k+1}} \cdot \left[ \frac{1}{x} f^{(k+1)}\left(\frac{1}{x} \right) + k f^{(k)} \left(\frac{1}{x} \right) \right] \\
&=  - \frac{(-1)^k}{x^{k+1}} \cdot \frac{1}{x} f^{(k+1)}\left(\frac{1}{x} \right) \\
&=  \frac{(-1)^{k+1}}{x^{k+2}} \cdot f^{(k+1)}\left(\frac{1}{x} \right)
\end{align*}
$\therefore P_k \implies P_{k+1} $
$\therefore P_0 \wedge (P_k \implies P_{k+1}) $
$\therefore P_n \quad \forall n \in \mathbb{W} $
