Prove: If $\alpha=((n+ \sqrt{n^2-4})/2)^m$ when $n\geqslant2$ and $m$ are natural numbers, prove that $\alpha=(k+\sqrt{k^2-4})/2$, $k\in\Bbb N$ 
Consider the number
$$
    \alpha=\left(\frac{n+ \sqrt{n^2-4}}2\right)^m
$$
where $n\geqslant2$ and $m$ are natural numbers.
Prove that
$$
    \alpha=\frac{k+\sqrt{k^2-4}}2
$$
for some $k\in\Bbb N$.
What I've tried:
I noticed that $\alpha$ can be seen as a root raised to the $m$-th power, for a quadratic expression $x^2 -nx +1$.  However, I am not sure how to leverage this to get to the required answer.
 A: $n=2$ is trivial.  For any integer $n>2$, let $\alpha_m = \displaystyle \left(\frac{n+\sqrt{n^2-4}}2\right)^m = \alpha_1^m$.
As $\alpha_1$ is a root of the quadratic $x^2-nx+1$, we can substitute $\alpha_1$ and multiply throughout by $\alpha_1^m$ to get that $\alpha_i$ satisfies the recursion $\alpha_{m+2}-n\,\alpha_{m+1}+\alpha_m=0$ or $2\alpha_{m+2}=2n\,\alpha_{m+1}-2\alpha_m$.  It can be checked easily that $2\alpha_1, 2\alpha_2$ are of the form $a+b\sqrt{n^2-4}$ for integers $a, b$, hence the recursion guarantees $\alpha_m$ is also of desired form.
A: For $n=2$ is obvious. Let $n>2$ integer. Then the equation
$$
x^2-nx+1=0
$$
has two roots $r,s$, both real, satisfying
$$
0<r=\frac{n-\sqrt{n^2-4}}{2}<1<s=\frac{n+\sqrt{n^2-4}}{2}
$$
We shall show inductively, in $k$, that $r^k,s^k$ are the roots of $x^2-n_kx+1=$, for a suitable $n_k>2$ positive integer.
For $k=1$ is already known. Also, $r^ks^k=1$ for all $k\in\mathbb N$.
Assume that for some $k$, we have $r^j+s^j=n_j\in\mathbb N$, for all $j=1,\ldots,k$.  Then
$$
n\cdot n_k=(r+s)(r^k+s^k)=r^{k+1}+s^{k+1}+rs(r^{k-1}+s^{k-1})
=(r+s)(r^k+s^k)=r^{k+1}+s^{k+1}+(r^{k-1}+s^{k-1})
$$
and hence
$$
r^{k+1}+s^{k+1}=n\cdot n_k-(r^{k-1}+s^{k-1})=n\cdot n_k-n_{k-1}.
$$
Clearly $n_{k+1}=r^{k+1}+s^{k+1}$ is an integer, and it is positive since $r,s>0$. Also, $r^{k+1},s^{k+1}$ are clearly the roots of $x^2-n_{k+1}x+1$, because their product is 1 and their sum is $n_{k+1}$.
