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Show that $$\mathbb{Z}[\sqrt{5}]^{\times} = \{\pm(2 + \sqrt{5})^k : k \in \mathbb{Z}\}$$

The inclusion $\supseteq$ is fairly easy by concerning that $\mathbb{Z}[\sqrt{5}]$ is a multiplicative group and the norm function. The inclusion which I have trouble with is $\subseteq$. In the solutions of my book we assume at the beginning, that if $x \in \mathbb{Z}[\sqrt{5}]$, then we can assume $x \geq 1$ since $\pm x,\pm x^{-1} \in \mathbb{Z}[\sqrt{5}]$. I do not fully understand this. Furthermore, there exists $m \in \mathbb{N}$ such that $$(2 + \sqrt{5})^m \leq x < (2 + \sqrt{5})^{m + 1}$$ which I also do not fully understand. Any help or any suggestions for a different proof would be appreciated.

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    $x$ is in particular a real number. If $x<0$ consider $-x>0$. Now can assume $x>0$. If $01$. – 2017-01-15
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    The powers of $2+\sqrt5$ form a strictly ascending chain of real numbers $\ge1$ (for $m=0$ you get $1$). Then your $x\ge 1$ must fall between two consecutive elements of this sequence since they split the interval $[1,\infty)$ into disjoint intervals. – 2017-01-15
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    There are many helpful answers for this on MSE, see for example [here](http://math.stackexchange.com/questions/124775/find-units-in-mathbbz-sqrtn) and the link to related questions. – 2017-01-15

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