Show that $(\sqrt{2}-1)^n$ is irrational 
Show that for all $n\in \mathbb{N}$ the number $(\sqrt{2}-1)^n$ is irrational.

I do not get the idea of the proof at all, any help appreaciated.
edit: I am also thinking whether it will be possible to show $(\sqrt{2}-1)^n=\sqrt{m+1}-\sqrt{m}$
 A: Since 
$$
(\sqrt{2}-1)^n=\frac{1}{(\sqrt{2}+1)^n}
$$
it is enougth to prove that $(\sqrt{2}+1)^n$ is irrrational.
We have after direct calculation  that 
$$
(\sqrt{2}+1)^n=a+b \sqrt{2}, 
$$
for some  natural   $a,b,$  $b\neq 0.$
Suppose now that $(\sqrt{2}+1)^n$ is rational. It  follows  that 
$\sqrt{2}$ is also rational since 
$$
\sqrt{2}=\frac{1}{b}((\sqrt{2}+1)^n-a).
$$
We obtain  a contradiction so 
 $(\sqrt{2}-1)^n$ is an irrational number.
A: Note that if we set $a=1-\sqrt 2$ and $b=1+\sqrt 2$ we have $a+b=2$ and $ab=-1$, so that $a,b$ are roots of the quadratic $x^2-2x-1=0$, and $(ab)^n=(-1)^n$.
Now note that $a^2-2a-1=0$ can be multiplied by $a^n$ to obtain the equation $a^{n+2}=2a^{n+1}+a^n$. If we add the equivalent equation for $b^{n+2}$ and set $u_n=a^n+b^n$ we find that $$u_{n+2}=2u_{n+1}+u_n$$
Since $u_0=u_1=2$ the $u_n$ are increasing positive (even) integers and $a^n$ and $b^n$ are roots of $x^2-u_nx+(-1)^n=0$.
Now apply the rational root theorem.
A: $(\sqrt{2}-1)^n$ can be written as $p(n) + q(n)\sqrt{2}$, with both $p(n)$ and $q(n)$ integers, with $q(n)\neq0$.
A: Let a vector $[x\ y]^T$, $x,y \in \mathbb{Q}$ represent a number of the form $x + \sqrt{2}y$. Multiplying by $(\sqrt{2} - 1)$ then corresponds to applying the matrix
$$ A = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}. $$
Multiplying by $(\sqrt{2} - 1)^n$ corresponds to applying $A^n$.
Suppose $(\sqrt{2} - 1)^n = r \in \mathbb{Q}$ for some $n \in \mathbb{Z}^+$. That would mean that $A^n = rI$. We show that it is impossible.
$A$, considered as a real matrix, has two distinct eigenvalues: $\pm\sqrt{2} - 1$. Thus, $A^n$ also has two distinct eigenvalues: $(\pm\sqrt{2} - 1)^n$. But then it can't be of the form $rI$.
