Differential equation involving nth derivative and inverse functions. So I recently came across a video involving function inverses and nth derivatives, and in the comments I saw the following result.
Given function $f(x) = ax^b$, the solution to the differential equation $f^{(n)}(x) = f^{-1}(x)$ came out to $a={n+\sqrt{n^2+4}\over 2}$ and $b={\Gamma(a-n+1)\over \Gamma(a+1)}^\frac{1}{a-n+1}$. 
Then solution of $\lim_{n\to 0}f^{-1}(x)=f(x)$ appears numerically to approach $e^{1-\gamma}$ 
The original poster and I both have no idea why.
Is there a proof or intuitive explanation that can show this result?
The original calculations:https://www.desmos.com/calculator/zzui7ezcho
 A: If $\;f(x) := b\,x^a\,$ (as in your link!) then $\;f^{(n)}(x) = b\,x^{a-n}\,a(a-1)(a-2)\cdots (a-k+1)\;$ that is :
$\displaystyle f^{(n)}(x) = b\,x^{a-n}\frac{\Gamma(a+1)}{\Gamma(a-n+1)}\ \ $ for $n$ a non-negative integer.
Supposing the fractional derivative chosen so that this rule generalizes to non integer values of $n$ ; $f^{(n)}(x) = f^{-1}(x)\ $ then becomes :
$$b\,x^{a-n}\frac{\Gamma(a+1)}{\Gamma(a-n+1)}=\left(\frac xb\right)^{\large{\frac 1a}}$$
this will be true for any positive $x$ for $\;\displaystyle a=\frac{n\pm\sqrt{n^2+4}}2,\;b=\left(\frac{\Gamma(a-n+1)}{\Gamma(a+1)}\right)^{\large{\frac 1{1+1/a}}}$.
As $\,n\to 0\,$ we get the expansion for the positive solution $a=a(n)$ :
\begin{align}
a(n)&=1+\frac n2+O(n^2)\\
\text{that we will use for $b(n)$ :}\qquad\\
b(n)&=\left(\frac{\Gamma\left(2-\frac n2+O(n^2)\right)}{\Gamma\left(2+\frac n2+O(n^2)\right)}\right)^{\frac 12+O(n)}\\
b(n)&=\exp\left(\left(\ln\Gamma\left(2-\frac n2+O(n^2)\right)-\ln\Gamma\left(2+\frac n2+O(n^2)\right)\right)\left(\frac 12+O(n)\right)\right)\\
\text{using}\ &\ln\Gamma(2+z)=\ln\Gamma(1+z)+\ln(1+z)=(1-\gamma)z +O(z^2)\\
\text{from}\ \,&\ln \Gamma(1+z)= -\gamma z +\sum_{k=2}^\infty \frac{\zeta(k)}k\,(-z)^k\;\text{ for}\;|z| < 1\quad\text{we obtain}\\
b(n)&=1+(\gamma-1)\frac n2+O(n^2)\\
\end{align}
The positive solution $x$ of $\;\displaystyle b\,x^a=\left(\frac xb\right)^{\large{\frac 1a}}\;$ or $\;\displaystyle b^{1+a}=x^{1-a^2}\;$will thus be given by :
$$x=\lim_{n\to 0} \;b(n)^{\large{\frac 1{1-a(n)}}}=\lim_{n\to 0} \;\left(1+(\gamma-1)\frac n2+O(n^2)\right)^{\large{-\frac 2{n+O(n^2)}}}$$
Which is the conjectured $\;\displaystyle x=e^{\large{1-\gamma}}$.
