Golden Ratio Approximation $$\sqrt{1000}-30.0047 \approx \varphi $$
$$[(\sqrt{1000}-30.0047)^2-(\sqrt{1000}-30.0047)]^{5050.3535}\approx \varphi $$
Simplifying Above expression we get
$$1.0000952872327798^{5050.3535}=1.1618033..... $$
Is this really true that 
$$[\varphi^2-\varphi]^{5050.3535}=\varphi $$
 A: No, not at all. Factoring and then using a well known property of $\varphi$: $$\left(\varphi(\varphi-1)\right)^{5050.3535}=\left(\frac{\varphi}{\varphi}\right)^{5050.3535}=1\ne\varphi. $$
The argument needs to be a bit larger than $1$. You're dealing with an approximation.
A: This is not an answer but it is too long for a comment.
Working with illimited precision, let $$a=\sqrt{1000}-\frac{300047}{10000}\approx 1.6180766016837933200$$ $$(a^2-a)^{\frac {50503535}{10000}}\approx 1.6180331121536741389$$ $$\phi\approx 1.6180339887498948482$$ Using as exponent $5050.3592$ (same number of digits as in your post) instead (and doing the same), you would get  $$(a^2-a)^{\frac {50503592}{10000}}\approx 1.6180339909260630347$$ which is closer but still not exact (Vincenzo Oliva clearly explained the problem).
A: Of course, this is an approximation. No need for more proof than those already given by Vincenzo Oliva and Claude Leibovici - kindest regards !
I would add that it is easy to find a lot of such approximations.
Some amazing ones, to be compared to :
$1.6180339887...\simeq\frac{1+\sqrt5}{2}=\varphi$
$1.6180339886...\simeq\cos (\sqrt2\: e^{-2})-\cos(\sqrt[3]{2}\:e^\pi )$
$1.6180339884...\simeq\sqrt{\cosh(\gamma)+\cos(\gamma)}-\frac{\gamma^3}{\sin(5)}$ where $\gamma$ is the Euler constant.
$1.6180339881...\simeq\cosh(G^2\sinh(1))+\cos\left(\frac{\sqrt[3]\pi}{\cos(3)} \right)$ where $G$ is the Catalan constant.
They comes from : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales  (page 4)
This paper roughly explains the method to compute a lot of approximations of this kind (In French, presently no translation avalable).
