Show that $\bigcap_{n\in\mathbb{N}}=(0,1) $ if {$ {A_n}$}$_{n\in\mathbb{N}} =$ {$A_n$:$ A_n=(0, n^k); n,k \in\mathbb{N}$} [closed]

I'm using the property of $A=X\cap Y$ for n sets i.e.

$A=\bigcap_{ n\in\mathbb{N}}A_n \iff$(($\forall n\in\mathbb{N}, A\subseteq A_n$) and ($B \subseteq A_n ,\forall n \in \mathbb{N} \implies B \subseteq A$))

Now comes the solution part.

$ \forall n \in \mathbb{N}, n \geq 1\implies n^k \geq n \geq1$ as ($k \in \mathbb{N} \implies k \geq 1$)

$\therefore 0 \leq 1 \leq n^k$ i.e. $1$ is an interior point of each such interval except $(0,1)$

Hence using the corollaries of definition of intervals, we can conclude that $\forall n\in\mathbb{N}, A\subseteq A_n$

I'm a bit confused about the next step. Can anyone help me figure it out or suggest me a better approach to the question? Am I missing something while figuring out the solution?