$x\in \mathbb{R}\setminus\{0\}$ and $x + \frac{1}{x}$ is an integer. Prove $\forall n\in \mathbb{N}$ $x^n + \frac{1}{x^n}$ is also an integer. 
Let $x$ be a non-zero real number such that $x + \frac{1}{x}$ is an integer. Prove that $\forall n\in \mathbb{N}$ the number $x + \frac{1}{x}$ is also an integer.

Attempt at solution using induction:
base case: $n = 1$ then $x^n + \frac{1}{x^n} = x + \frac{1}{x}$ which is an integer
Inductive assumption: Assume that for some $k\in \mathbb{N} : x^k + \frac{1}{x^k}$
We have to show that $ x^{k+1} + \frac{1}{x^{k+1}}$ is an integer.
$$ x^{k+1} + \frac{1}{x^{k+1}} = x^k\cdot x + \frac{1}{x^k\cdot x} =  \frac{(x^k\cdot x)\cdot (x^k\cdot x)+1}{x^k\cdot x}$$
I can't find a way to seperate $x$ and $\frac{1}{x}$ from the term so I can use the inductive assumption.
Edit: I don't think that this question should count as a duplicate since the linked question is asking to specifically solve another problem and one of the answers of that question utilize the proof from this question in their answer but that question itself is different and not related to this question since that question can be solved without using this proof as other answers of that question don't include it.

Both, $x$ and $\dfrac 1x$ are roots of the same equation $X^2-aX+1=0$ 
where $a$ is an integer. It follows that any equation deduced from it is also an equation of both  $x$ and $\dfrac 1x$.
We have 
$$X^2=aX-1$$ $$X^3=aX^2-X=a(aX-1)-X=(a^2-1)X-a$$
 $$X^4=(a^2-1)X^2-aX=(a^2-1)(aX-1)-aX=(a^3-2a)X-(a^2-1)$$ 
For $X^n$ one has by iteration$$X^n=f_n(a)X+g_n(a)$$ where $f_n(a)$ and $g_n(a)$ are integers.
Since also $$\left(\frac{1}{X}\right)^n=f_n(a)\left(\frac{1}{X}\right)+g_n(a)$$ we conclude that
$$x^n+\frac{1}{x^n}=f_n(a)(x+\frac1x)+2g_n(a)=af_n(a)+2g_n(a)\in\mathbb Z$$
(Note that this mode allows us to calculate the integer values of  $x^n+\frac{1}{x^n}$).
