Show that for a principal ideal in $O_K$ we have $(\alpha)=(\varepsilon^k \alpha)$ where $\varepsilon$ is the fundamental unit

Let $K$ be a real quadratic number field and $O_K$ it's number ring. Let $\varepsilon$ be the fundamental unit of $O_K$. Given $\alpha \in O_K$ I want to show that the principal ideal generated by $\alpha$ is the same as the ideal generated by $\varepsilon^k \alpha$ for some $k\in \mathbb{Z}$. I.e.: $$ (\alpha)=(\varepsilon^k\alpha). $$ Clearly we have $(\varepsilon^k\alpha)\subseteq (\alpha)$ as $\alpha \mid \varepsilon^k\alpha$, but how does one show that $(\varepsilon^k\alpha)\supseteq (\alpha)$?

(This is a small result one need to understand to prove Dirichlet's class number theorem. See exc. 10.5.9 in "Problems in Algebraic Number Theory" M. Ram Murty; Jody Esmonde.)

$\textbf{Fundamental unit:}$ When $K$ is real quadratic number field there exist a unit $\varepsilon$ s.t. $$ U=\{\pm \varepsilon ^k | k\in \mathbb{Z}\} $$ where $U$ is the unit group of $O_K$. We call $\varepsilon$ the fundamental unit of $O_K$. Further we know that $U \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$, again only when $K$ is real quadratic.

As @Mathmo123 noted; above holds for any unit. We have by properties by ideals that for any $r\in O_K$ we have $r(\alpha)\subseteq (\alpha)$. As for any unit $u\in O_K$ we know that $u^{-1}\in O_K$, hence $(\alpha)=u^{-1}(u\alpha)\subseteq (u \alpha)$, which is what we wanted.